## New answers tagged ac.commutative-algebra

4
votes

Accepted

### An assertion of Mahler

Let us introduce the notation
$$R:=\mathbb{C}[x_1,\dotsc,x_{g-1}]\qquad\text{and}\qquad K:=\mathbb{C}(x_1,\dotsc,x_{g-1}).$$
Let us regard
$$F:=F(x_1,\dotsc,x_g)\qquad\text{and}\qquad G:=F(x_1^\rho,\...

0
votes

### Krull dimension of completions in non-noetherian setting (especially completed perfections)

It turns out that the rings you mention, often called "perfectoid Tate algebras" in the literature, indeed have infinite Krull dimension. In fact, the Krull dimension is uncountable in the ...

0
votes

### Deminormal and Gorenstein

If a deminormal scheme is irreducible, it is normal automatically since a node point (of codimension 1) cannot happen and that means your scheme is already of $R_1$ and $S_2$ (cf. Serre's criterion ...

1
vote

### Basic question about completion of local ring

So I also think (iii) is true.
Claim: if $A$ is a Noetherian local ring and its $\mathfrak m$-adic completion $A^\wedge$ is essentially of finite type over $A$, then $A$ is a complete local ring.
...

1
vote

### What is interesting/useful about big Witt Vectors?

Other answers addressed the Witt vector functor as a right adjoint. However, it is interesting that, in equivariant homotopy theory, it can also be realized as a left adjoint.
More precisely, let $G$ ...

5
votes

Accepted

### Dimension from Hilbert series with variable-weighted grading?

What you have to do is look at the pole at $t=1$, and its order gives you the Krull dimension. This is discussed in detail in Proposition 5.3.2 in my book, "Representations and Cohomology II"...

0
votes

### The inverse limit of a sequence of ring surjections commutes with taking difference subsets of the respective units & gluing in some primes?

In general let the $A_i$ be any subsets of $\Bbb{Z}/p_i\#$ such that all together they form an inverse-system, with maps that are restrictions of ring homomorphisms, respectively to each $A_i$.
Then $$...

12
votes

Accepted

### If $E_\text{sep}/F$ is normal, then must $E/F$ be normal?

Let $K=M(X,Y)$ be the function field in two variables, for a field $M$ of characteristic $p>0$.
Consider the Galois homomorphism $\gamma: f(X,Y)\mapsto f(Y,X)$ and let $F=K^\gamma$ the fixed field. ...

4
votes

### Randomly fixing elements and transcendence degree

Let me give a partial answer ignoring issues of inseparability. The map $\mathbb A^{m-j}_{\mathbb F_q(x_1,\dots,x_j)} \to \mathbb A^n_{\mathbb F_q(x_1,\dots,x_j)}$ has image of dimension $r$. If $\...

6
votes

### Basic question about completion of local ring

Here are short proofs of (i) and (ii).
(i) If $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ then the completion $\hat A$ is a local ring with maximal ideal $\hat{\mathfrak{m}}=\...

1
vote

### Basic question about completion of local ring

Please let me know if this is wrong. I believe the answer to all three questions is yes.
Let $ (A, \mathfrak{m}) $ be a Noetherian, local ring and let $ \widehat{A} = \varprojlim_{i \in \mathbb{N}} A/...

0
votes

### Computing the minimal polynomial of roots of polynomials with algebraic coefficients

One approach would be as follows: Let $c_i^{(j)},\,j=1,\ldots,\deg q_i$ be the conjugates of $c_i$. Denote $p_{j_0,\ldots,j_n}(x)=\sum_{i=0}^nc_i^{(j_i)}x^i$ and form the product $P(x)=\prod_{j_1,\...

5
votes

### Nonzero module with vanishing derived fibers

The following example is from Sasha Petrov (any mistakes are mine)
Let $S$ be the ring $k[x^{1/p^\infty}]$ for some field $k$, and $N$ be the maximal ideal $(x^{1/p^\infty})$ in $S$. Then take $R=S/x$ ...

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1
vote

Accepted

### An example of a commutative ring $R$ which has a proper right ideal which is not a right SIP $R$-module

Now, I ask if there exists a commutative ring 𝑅 and a proper right ideal $A\subset R$ such that the module $A_R$ is not an $\textsf{SIP}$.
Yes. Start with any commutative ring $S$ that has a non-$\...

5
votes

Accepted

### Idempotent algebras over absolutely flat ring

They must all be discrete.
Let $k$ be absolutely flat - for any $k$-modules $M,N$, we have $\pi_*(M\otimes_k N)= \pi_*(M)\otimes_k \pi_*(N)$. Now if $A$ is idempotent over $k$, let us examine what ...

6
votes

### Reference for surjectivity of the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$

I think that statement is false.
Let $R=\mathbb{Z}[\varepsilon]/\varepsilon^2$ and let $G_1=\mathbb{Z}/2\times 1$ and $G_2=1\times \mathbb{Z}/2$ and $G=\mathbb{Z}/2\times\mathbb{Z}/2$. Let each $G_i$ ...

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