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 $R=S/x$ ...
Community wiki
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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