show/hide this revision's text 3 Improve grammar

I reject the premise of the question. :-)

It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears as part of in Matthew Emerton's proof. As Terry says, what is really nice about this proof is that nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finitely-dimensional finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.

The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relatorsrelations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give in the rest of this answernext.

Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.

Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$. x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )

The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.

The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A = \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □

People should compare

Compare the complexity of this argument to the other arguments supplied so far.

Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ actually divides every coefficient in the presentation, so that and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.

Another way to make describe the point (besides that arguing finitely many relations is not important) argument is that Smith normal form is nothing other than a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.

The proof immediately works for any Euclidean domain, ; in particular, it also implies the Jordan canonical form theorem, and . And it you only need needs minor changes to make it a little more abstract apply to make it work for general PIDs.
show/hide this revision's text 2 Rm shallow point of confusion about Matthew Emterton's proof

I reject the premise of the question. :-)

It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears as part of Matthew Emerton's proof. As Terry says, what is really nice about this proof is that it works finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finitely-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.In Matthew's proof, it doesn't seem obvious to me that $A_{\mathrm{tors}}$ is finitely generated just because $A$ is finitely generated.

The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relators. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give in the rest of this answer.

Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.

Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )

The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.

The coefficient $z$ of $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A = \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □

People should compare the complexity of this argument to other arguments supplied so far.

Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ actually divides every coefficient in the presentation, so that each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.

Another way to make the point (besides that arguing finitely many relations is not important) is that Smith normal form is nothing other than a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.

The proof immediately works for any Euclidean domain, in particular it also implies the Jordan canonical form theorem, and it you only need to make it a little more abstract to make it work for PIDs.
show/hide this revision's text 1

I reject the premise of the question. :-)

It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears as part of Matthew Emerton's proof. As Terry says, what is really nice about this proof is that it works finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finitely-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question. In Matthew's proof, it doesn't seem obvious to me that $A_{\mathrm{tors}}$ is finitely generated just because $A$ is finitely generated.

The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relators. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give in the rest of this answer.

Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.

Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )

The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.

The coefficient $z$ of $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A = \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □

People should compare the complexity of this argument to other arguments supplied so far.

Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ actually divides every coefficient in the presentation, so that each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.

Another way to make the point (besides that arguing finitely many relations is not important) is that Smith normal form is nothing other than a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.

The proof immediately works for any Euclidean domain, in particular it also implies the Jordan canonical form theorem, and it you only need to make it a little more abstract to make it work for PIDs.