## New answers tagged ac.commutative-algebra

11

The necessary and sufficient condition is that $J$ be generated by an idempotent.
A. Assume $J=(e)$ is generated by an idempotent $e$ ($e^2=e$). Then the annihilator of $J$ contains $1-e$, so the sum of $J$ and its annihilator contains $e + (1-e)=1$. This shows that $J$ and its annihilator are comaximal.
B. Now suppose that the annihilator $K$ of $J$ is ...

0

This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any nonzero prime of $D$, so $S = D \setminus 0$. Thus $a \in S$ which means some power of $b$ divides $a$.
If you had an ascending chain $(a_i)$ which satisfied, ...

ag.algebraic-geometry ac.commutative-algebra algebraic-number-theory dedekind-domains valuation-rings

1

Yes. This is done essentially as you tried by taking the inverse images of $I$ in the $R_i$ under the canonical morphisms. For a detailed explanation see EGA 0.6.1.2 (in the 1971 Springer edition).

2

The proof of (LC3), in the given setting, is surprisingly difficult, or, at least, elaborate. Let $a \in A$ with $\lambda_S(a) = [a] = 0$ in $A[S^{-1}]$,
i.e. one has
\begin{equation}
\tag{1}
a \in (sT_s-1\,|\,s \in S).
\end{equation}
To show is that
\begin{equation}
\tag{2}
sa = 0
\end{equation}
for some $s \in S$. Because of (1), there are elements ...

2

Put $\mathcal{A}'=\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and similarly for $\mathcal{D}'$. Choose an isomorphism $f\colon\mathcal{D}'\to\mathcal{A'}$.
Choose a (countable) basis $\mathcal{D}_0=\{d_i:i\in\mathbb{N}\}$ for $\mathcal{D}$ over $\mathbb{C}$. Then $f(\mathcal{D}_0)$ is a basis for $\mathcal{A}'$ over $\mathbb{C}(t)$ but $\dim_{\mathbb{C}}(...

5

For finitely generated domains over a base field $k$ of characteristic 0, we have that if $A$ is regular, then both $Der_k \, A$ and the module of Khäler differentials are finitely generated projective (McConnell, Robinson, Noncommutative Noetherian Rings, revised edition, 15.2.11).
Zariski-Lipman's Conjecture says that if $Der_k \, A$ is finitely generated ...

0

This is still is equivalent to JC.
Your equality says, $\mathbb{C}[x,y]=\mathbb{C}[f,g]+g\mathbb{C}[x,y]$, the last term is equal to $\mathbb{C}[f]+g\mathbb{C}[f,g]+g\mathbb{C}[x,y]=\mathbb{C}[f]+g\mathbb{C}[x,y]$, since $g\mathbb{C}[f,g]\subset g\mathbb{C}[x,y]$. This says, the map $\mathbb{C}[f]\to \mathbb{C}[x,y]/g\mathbb{C}[x,y]$ is onto and then it is ...

0

Let $R=\mathbb{C}[f_1,\ldots,f_n]\subset\mathbb{C}[x_1,\ldots, x_n]=A$. A derivation $U$ of $A$ is completely determined by $U(x_i)$, since, then $U(P(x_1,\ldots,x_n))=\sum\frac{\partial P}{\partial x_i} U(x_i)$ for any $P\in A$. In particular, one has $U(f_i)=\sum \frac{\partial f_i}{\partial x_j}U(x_j)$. If $U_{|R}=D$, then $U(f_i)=D(f_i)$. Now using the ...

6

Here is a proof that $$\ker \lambda = \{ a \in A \, \vert \, sa = 0 \text{ for some } s \in S\} \quad (LC_3)$$ holds true assuming that the following definition is in use: $$A[S^{-1}] = A[T_s \vert s \in S] /\left(sT_s -1 \vert s \in S\right).$$
If $ta = 0$ for some $a \in A$ and some $t \in S$, then we have $$a = -(tT_t - 1)a \in (sT_s -1 \vert s \in S).$$ ...

14

While it may be, in general it is not. Consider as an example, $u=x+y+yp(x), v=x+yp(x)$ where$ \deg p(x)\geq 2$. Then $(u,v)=(x,y)$ and so maximal. Notice that $u=y+v$. So, $$(u-a, v-b)=(u-v+b-a, v-b)=(y+b-a, v-b)=(y+b-a, x-b+(a-b)p(x))$$
and so most pairs of values of $a,b$, it is not maximal, as long as $\deg (x+(a-b)p(x))\geq 2$.

1

I just thought of a new proof that is for free using modern technology. It resembles the first step of the other proof, but replacing the sheaf of differentials with the cotangent complex.
In order to see that the map $f:R\to S$ is formally étale, it will be enough to show that the entire cotangent complex vanishes. The property of a (derived) module being ...

3

Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice.
Take
$$
A=
\begin{pmatrix}
0&1&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{pmatrix},
$$
$$...

3

Assuming that you are asking for an isomorphism taking $B$ to $B$':
No. Take $A$ to be the $n\times n$ Jordan block with eigenvalue zero. The centralizer of $A$ is exactly $\mathbb C[A]$. Hence for any $B,B' \in \mathbb C[A]$, we have $\mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']$. However, $B$ and $B'$ need not be conjugate: take $B = A^2, B' = A^3$. As ...

8

The following works over any ring $R$: Take a family $\underline{X}:=(X_a)_{a\in R}$ of indeterminates indexed by $R$, and put $R_1:=R[\underline{X}]/I$ where $I$ is generated by $(X_a^2-a)_{a\in R}$. Then $R_1$ is free as an $R$-module (you can view it as $\bigotimes_{a\in R}R[X_a]/(X_a^2-a)$) and every element of $R$ becomes a square in $R_1$. Now just ...

answered Jul 16 at 13:50

Laurent Moret-Bailly

16.6k11 gold badge5151 silver badges7373 bronze badges

3

In general, $(P, X)$ is not the only prime containing $P[[X]]$ and contracting to $P$. I don't have anything to say about the problem of characterizing such primes, but in general it seems extremely hard. Let's focus on the case $P = 0$.
As a motivating example we can even use the integers. The ring $\mathbb{Z}[[X]]$ is a UFD. For any prime $p$ and power ...

7

$\def\CC{\mathbb{C}}$User "anon" points out to me that this is Proposition 8.28 in Milne's notes; see also Example 8.36 for a quasi-finite map $\CC^2 \to \CC^2$ which is not finite. The rest of my answer is probably not as useful now that there is a good reference, but I'll leave it.
Here is the right statement:
Theorem Let $X$ and $Y$ be affine ...

answered Jul 15 at 15:02

David E Speyer

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4

This is not true even for affine schemes. Let $k = \mathbb{Z}$, let $X = \operatorname{Spec} \mathbb{Z}$, let $Y = \operatorname{Spec} \mathbb{F}_p$, and let $Z \cong \operatorname{Spec} \mathbb{Z} [ p^{-1}]$. The closure of $Z$ in $X$ is $X$ itself, but $Z \times Y \cong \operatorname{Spec} \{ 0 \}$, which is already closed in $X \times Y$. Of course, $Y$ ...

0

### Infinite uniform dimension $\Rightarrow$ infinitely many idempotents in a localization of a quotient

In general, it is not possible to produce such an ideal $I$ and element $r \in R \setminus I$ from your assumption.
An apparent obstacle is the class of rings of infinite uniform dimension having exactly one prime ideal. If a ring $R$ has a unique prime ideal, then the same is true of every nonzero quotient and localization. In particular, $R$ doesn't have ...

3

We have
$$\binom{n}d_q=\frac{(q^n-1)\ldots(q^n-q^{d-1})}{(q^d-1)\ldots(q^d-q^{d-1})}=f_d(q^n)=g_d([n]_q)
$$
where $f_d$ and $g_d$ are polynomials of degree $d$ (depending on $q$ of course). Therefore $$\bigl( P([0]_q), P([1]_q), \ldots, P([N-1]_q) \bigr) \in \bigl\langle u^{(0)}_q, u^{(1)}_q, \ldots, u^{(d)}_q \bigr\rangle$$
if and only if $\deg P\leqslant d$...

5

Denote $a-b=x$, then $a^2-b^2=x(x+2b)=4z$ for $z\in R$. Assuming that $R$ is integrally closed, we see that $(x/2)^2+b(x/2)-z=0$, so $x/2$ is an algebraic integer, thus $x/2\in R$.

4

The "midpoint" operation $(x,y)\mapsto\frac12(x+y)$ on the reals is commutative and not associative.

2

To see what's going on, let's describe the free commutative, non-associative, algebra on $n$ elements $x_1, \dots, x_n$, which we call $A_n$ .
Let $X_n$ be the set of isomorphism classes of binary rooted trees $T$ with a function $f: {\rm Leaves}(T) \to \{x_1, \dots, x_n\}$, which labels the leaves. Then $A_n = k X_n$ is the free vector space with basis $...

7

The class of Jordan algebras are the most important class of algebras in this direction.
They are defined by the two identities,
(commutativity): $xy=yx$,
(Jordan identity): $(xy)(xx)=x(y(xx))$.
They are an extremely important class of non-associative algebras (check Jacobson's Strucuture and Representation of Jordan algebras or McCrimmon's A taste of Jordan ...

22

If you set $T=0$ or $X=0$ then you get the series $\sum_{n\geq 0} Y^{q^n}$. This cannot be rational because a rational power series in one variable that is not a polynomial cannot have arbitrarily long sequences of 0 coefficients (since the coefficients satisfy a linear recurrence relation with constant coefficients).

answered Jul 10 at 21:17

Richard Stanley

36.4k99 gold badges127127 silver badges205205 bronze badges

5

The answer is no.
Take $p=\frac{x^2}{2}$, take $q=\frac{y}{x}$. The Jacobian matrix is $\begin{pmatrix} x& -\frac{y}{x^2} \\ 0 & \frac{1}{x}\end{pmatrix}$ whose determinant is equal to $1$. However, $f$ is definitely not an automorphism of $\mathbb{C}(x,y)$.
More generally, take any polynomial $p\in\mathbb{C}[x]$ and choose $q=\frac{y}{p_x}$. This ...

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