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edited Jul 26 2010 at 19:34 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint- leaving . As a result, many students often struggling with alternative dense cannot easily resist the obvious topological approachestemptations and instead derive hairier proofs employing density arguments (e.g see elswhere in this thread).
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$
$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$
Now, for our pièce de résistance (of limits, density...)topology, we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\;\rm det \:A\;$ in all of the examples given above.
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edited Jul 18 2010 at 19:00 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$
$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$
Now, for our pièce de résistance (of limits, density...), we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm \;\rm det \:A$ :A\;$ in all of the example sparking this threadexamples given above.
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edited Jul 18 2010 at 19:00 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$
$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$
Now, for our pièce de résistance (of limits, density...), we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\;\rm \rm det \:A\;$ :A$ in all of the examples given aboveexample sparking this thread.
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edited Jul 18 2010 at 18:38 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$
$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$
Now, for our pièce de résistance (of limits, density...), we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm \;\rm det \:A$ :A\;$ in all of the example sparking this threadexamples given above.
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edited Jul 18 2010 at 18:22 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d d\: B = d d\: C$ for $\rm d = det det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d d\: b_{i,j} = d d\: c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ Z}[a_{i,j}\:]$ where $\;\rm d = det det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\rm \;\rm b_{i,j} = c_{i,j}$ c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}b_{i,j}\:, \; c_{i,j}\;$ c_{i,j}\:,\; d\;$ have polynomial form in $\;\rm a_{i,j}$a_{i,j}\:$, i.e. they are elt's elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\quad det(adj A\rm\quad\; det(I-AB) = (det A)^{n-1}\quad\;$ det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj A) = \;\;\rm (det A) \;I\quad\;\;$ I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det A$\:A$
$\rm\quad\; det(I-AB) \rm\quad\quad det(adj \:A) = det(I-BA)\;$ (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A \;\rm\quad A\;(adj\: A) = A\;(I-BA)\;$ (det\: A) \;I\quad\;\;$ then canceling $\;\rm det A$\:A$
Now. , for our pièce de résistance (of limits, density...), we derive the polynomial derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$$ \rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$
$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ :A$ in the example sparking this thread.
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edited Jul 18 2010 at 17:51 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad\quad det(adj A) = (det A)^{n-1}\quad\;$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj A) = (det A) \;I\quad\;\;$ then canceling $\;\rm det A$
$\rm\quad\; det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det A$
Now. for our pièce de résistance (of limits, density...), we derive the polynomial derivatives derivative purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad $ \rm f(x)g(x) - f(y)g(y) f(y)g(y)\; = (f(x)-f(y)) \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$g(x)-g(y))$$
$\rm\quad\quad\quad \quad\quad\quad\quad\rm\quad\quad\quad \implies Longrightarrow \quad\quad\quad\;\; quad\quad\quad\quad\quad\;
D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 17:34 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad \rm\quad\quad det(adj A) = (det A)^{n-1}\;$ A)^{n-1}\quad\;$ by taking $\;\rm det\;$ of $\;\rm \;\rm\quad A\;(adj A) = (det A) \;I\;$ ;I\quad\;\;$ then canceling $\;\rm det A$
$\rm\quad \rm\quad\; det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm \;\;\rm (I-AB)\;A = A\;(I-BA)$ A\;(I-BA)\;$ then canceling $\;\rm det A$
Now. for our pièce de résistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 17:27 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Now. for our piece pièce de resistance résistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 17:26 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Now, . for our pièce piece de résistanceresistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 15:34 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Now. , for our piece pièce de resistance résistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specialize the difference "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 15:25 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Now. for our pièce piece de résistanceresistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by dividing the first equation
by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$,
i.e. specializing specialize the difference quotient "quotient" from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 15:25 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Now. for our piece pièce de resistance résistance (of limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Resisting limits again, we get the product rule rule for derivatives from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by dividing the first equation by $\rm x-y$, then evaluating at $\rm y = x$,
i.e. specializing the difference quotient from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 15:11 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Finally
Now. for our piece de resistance (of limits!)limits, density...), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$
and from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Finally, resisting Resisting limits once again, we formally prove Leibniz's get the product rule rule for derivatives : from the trivial difference product rule
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y)) $g(x)-g(y))$
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by dividing the first equation by $\rm x-y$, then evaluating at $\rm y = x$,
i.e. specializing the difference quotient from the product rule for differences.
Here the formal cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det A$ \;A$ in the example sparking this thread.
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edited Jul 18 2010 at 14:54 |
HINT $\;$ Work "generically", generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}\;$ by taking $\;\rm det\;$ of $\;\rm A\;(adj A) = (det A) \;I\;$ then canceling $\;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)\;$ by taking $\;\rm det\;$ of $\;\rm (I-AB)\;A = A\;(I-BA)$ then canceling $\;\rm det A$
Finally. for our piece de resistance (of limits!), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)$ f(x)-f(y)\;$ in $\rm R[x,y]$
\;\rm R[x,y],\;$
and from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Finally, resisting limits once again, we formally prove Leibniz's product rule rule for derivatives:
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y)) $
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by dividing the first equation by $\rm x-y$, then evaluating at $\rm y = x$,
i.e. specializing the difference quotient from the product rule for differences.
Here the formal cancellation cancelation of the factor $\;\rm x-y\;$ before evaluation at $\;\rm y = x\;$ is precisely analogous to the formal cancelation of $\rm det A$ in the example sparking this thread.
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edited Jul 18 2010 at 14:47 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
Analogously, the same generic method of proof works form for many other polynomial identities, e.g.
$\rm\quad det(adj A) = (det A)^{n-1}$ A)^{n-1}\;$ by taking $\rm det$ \;\rm det\;$ of $\;\rm A(adj A\;(adj A) = (det A) I\;$ \;I\;$ then canceling $\rm \;\rm det A$
$\rm\quad det(I-AB) = det(I-BA)$ det(I-BA)\;$ by taking $\rm det$ \;\rm det\;$ of $\rm \;\rm (I-AB)A I-AB)\;A = A(I-BA)$ A\;(I-BA)$ then canceling $\rm \;\rm det A$
Finally. for our piece de resistance (of limits!), we derive polynomial derivatives purely formally.
For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows
from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)$ in $\rm R[x,y]$
and from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition
since it is linear and it takes the same value on the basis monomials $\rm x^n$.
Finally, resisting limits once again, we formally prove Leibniz's product rule rule for derivatives:
$\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y)) $
$\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $
by dividing the first equation by $\rm x-y$, then evaluating at $\rm y = x$,
i.e. specializing the difference quotient from the product rule for differences.
Here the formal cancellation of the factor $\rm x-y$ \;\rm x-y\;$ before evaluation at $\rm \;\rm y = x$ in x\;$ is precisely analogous to the formal cancelation of $\rm det A$ in the example sparking this thread.
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edited Jul 18 2010 at 14:37 |
Analogously, the same generic method of proof works form many other identities, e.g. $\rm\quad det(adj A) = (det A)^{n-1}$ by taking $\rm det$ of $\;\rm A(adj A) = (det A) I\;$ then canceling $\rm det A$ $\rm\quad det(I-AB) = det(I-BA)$ by taking $\rm det$ of $\rm (I-AB)A = A(I-BA)$ then canceling $\rm det A$ Finally. for our piece de resistance (of limits!), we derive polynomial derivatives purely formally. For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)$ in $\rm R[x,y]$ and from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition since it is linear and it takes the same value on the basis monomials $\rm x^n$.Finally, resisting limits once again, we formally prove Leibniz's product rule rule for derivatives: $\rm\quad\quad\quad f(x)g(x) - f(y)g(y) = (f(x)-f(y)) g(x) + f(y) (g(x)-g(y)) $ $\rm\quad\quad\quad \implies \quad\quad\quad\;\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg) $ by dividing the first equation by $\rm x-y$, then evaluating at $\rm y = x$,i.e. specializing the difference quotient from the product rule for differences.Here the formal cancellation of the factor $\rm x-y$ before evaluation at $\rm y = x$ in is precisely analogous to the formal cancelation of $\rm det A$ in the example sparking this thread.
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answered Jul 18 2010 at 6:42 |
HINT $\;$ Work "generically", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be indeterminates and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}]. \;$ We wish to prove $\rm B = C$ from $\rm d B = d C$ for $\rm d = det A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d b_{i,j} = d c_{i,j}$ in the domain $\rm R = {\mathbb Z}[a_{i,j}]$ where $\;\rm d = det A \ne 0$, so $\rm d$ is cancelable, yielding $\rm b_{i,j} = c_{i,j}$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the universality of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$.
Notice that the crucial insight is that $\;\rm d,\; b_{i,j}, \; c_{i,j}\;$ have polynomial form in $\;\rm a_{i,j}$, i.e. they are elt's of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}]$ which, being a domain, enjoys cancelation of elts $\ne 0$.
Working generically allows us to cancel $\rm d$ and deduce the identity before any evaluation where $\rm d\mapsto 0.$
Such proofs by way of universal polynomial identities emphasize the power of the abstraction of a formal polynomial (vs. polynomial function). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint - leaving students often struggling with alternative dense topological approaches.
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