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Questions tagged [commutative-algebra]

Questions about commutative rings, their ideals, and their modules.

Commutative algebra is the area of mathematics that deals with commutative rings and their ideals, as well as modules over commutative rings.

Many results and tools of commutative algebra are cornerstones of algebraic geometry. Important tools of commutative algebra include localization and completion of rings and modules.

16857 questions
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Relation between squarefree algorithms and radical of ideals

For univariate polynomials, I can compute the squarefree variant with the same roots as $p / \text{gcd}(p, dp/dx)$, which is simple enough. Then I learned about radical of ideals generated by multivariate polynomials, and it seems that it is…
lvella
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$Spec(R)$ Noetherian and going up theorem

Let $S \subseteq R$ be commutative rings with 1 and suppose $Spec(R)$ is a Noetherian topological space. How do we show that the number of each $T \in Spec(R)$ lying over $P \in Spec(S)$ is finite? I guess the idea is to use the going up theorem. We…
user6495
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Factoring maps between noetherian rings

Let $A,B$ be commutative noetherian rings, and let $f:A\to B$ be a ring map. Can one always factor $f$ as $A\to C\to B$ where $C$ is a noetherian ring, $A\to C$ is flat, and $C \to B$ is surjective?
flat comm
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$k \subset A \subset B$, $B\supset k$ f.g., $\text{codim}_k(A) < \infty$ $\Rightarrow$ $B \supset A$ f.g. module?

Does this hold? Let $k \subset A \subset B$ where $k$ is a field and $A,B$ are commutative rings. If $B$ is a finitely-generated ring over $k$ and $\dim_k(B/A) < \infty$ then $B$ is a finitely-generated $A$-module. I think the above is used in a…
Haderlump
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Formulation of Kähler differentials

I want to make sure my formulation of Kähler differentials is correct. Here it is. Please let me know if you see something that is incorrect. Let $A$ is a $k$-algebra via a ring homomorphism $f:k\to A$. A $k$-\textit{derivation}, or a derivation…
coolpenguin
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Decomposition of the Kähler differential module of a polynomial ring

This congruence is quoted from Matsumura, Commutative algebra. The explanation given for it is very brief and I believe I am missing some prerequisites to understand the proof. (26.J, Page 189) Let $k$ be a ring, $A$ a $k$-algebra,…
coolpenguin
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Kernel of morphism of rings $R[T_1, \dots, T_m] \to R[T_1, \dots, T_n]/(f_1, \dots, f_k)$

Consider a morphism surjective of rings $\varphi: R[T_1, \dots, T_m] \to R[T_1, \dots, T_n]/(f_1, \dots, f_k)$ such that $\varphi|_R = id_R$ (that is, a morphism of $R$-algebras). Is there an easy way to see that $\ker \varphi$ is finitely generated…
user480840
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Noether Normalisation Proof

I have the following version of Noether Normalisation Lemma: Let $R=F[y_1,\dots,y_n]$ be a finitely generated $F$-algebra. Then there exists a subset $\{x_1,\dots,x_k\}$ of $R$ which is algebraically independent over $F$ and such that $R$ is a…
Jason V
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Quotient isomorphic to tensor

Given a non zero element $x \in M$ and the nonzero submodule $M' = Ax$ of a commutative ring with unity $A$, and B a flat $A$-module, I'm trying to see that as I can show that $$ (\operatorname{Ann}(x))^e \subset m^e \subsetneq B $$ where $m$ is a…
Daniel Moraes
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Integral extension inside a polynomial ring over a field

Let $K$ be a field and $D = K[X]$. I need to show that if $f\in D$ is non constant, then the extension of rings $K[f]\subset D$ is integral, and if $A$ is a subring of $D$ which contains $K$ and has Krull dimension $1$, then $A$ is a finitely…
user82494
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Given ring morphisms $R \to A$, $R \to R'$, $\Omega_{A \otimes_R R'/R'} = \Omega_{A/R} \otimes_A (A \otimes_R R')$

Let $R \to A$ be a morphism of rings, making $A$ into an $R$-algebra. Denote the module of differentials of $A$ over $R$ by $\Omega_{A/R}$, with exterior derivative $d_{A/R}: A \to \Omega_{A/R}$. Let $R \to R'$ be a ring homomorphism and denote $A'…
user480840
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Kernel of morphism from polynomial ring over field, to some ring

In the book "Undergraduate Commutative Algebra" by Reid, there appears the following supposedly very easy exercise 0.5: Why isn't the ring $A$ of diagonal $2\times 2$ matrices over (for instance) $\mathbb{C}$, with $\alpha$ the matrix $\text{diag}…
Chen Van Dam
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Trouble understanding the definition of irrelevant ideal

In mathematics, the irrelevant ideal is the ideal of a graded ring generated by the homogeneous elements of degree greater than zero. More generally, a homogeneous ideal of a graded ring is called an irrelevant ideal if its radical contains the…
jk001
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Does there exist a non-zero module over $R$ which vanishes modulo a nilpotent ideal?

Let $R$ be a noetherian ring, $I$ a nilpotent ideal in $R$, and $M$ a module over $R$, of infinite type, such that $M/IM = 0$. Is it necessarily the case that $M=0$? If $M$ were of finite type, then we would immediately get a positive answer by…
user125639
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$V(f)\cup V(g) = V((f)(g))=V((fg))=V(fg)$?

$f,g$ are all in $k[x_1,...,x_n]. k$ is a field. I find the title confusing: for example $f=(x_1-a_1)...(x_n-a_n), g=(x_1-b_1)...(x_n-b_n)$. Then $V(f)\cup V(g)$ is a two-point set. But $V(fg)$ clearly has more than two points e.g.…
Jun Xu
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