Articles of commutative algebra

Support of $\operatorname{Hom}(R/I, M)$

Let $R$ be a Noetherian ring, $I$ be an ideal of $R$ and $M$ be an $R$-module. Is the following formula true? $\operatorname{Supp}\operatorname{Hom}(R/I, M)=\operatorname{Supp}(M) \cap V(I)$ If not, under what conditions the above formula is true? Thanks.

Constructing an example s.t. $\operatorname{Hom}_R(M,N)$ is not finitely generated

This question already has an answer here: When $\operatorname{Hom}_{R}(M,N)$ is finitely generated as $\mathbb Z$-module or $R$-module? 1 answer

How is this proof correct in regard to a $k$-subalgebra (Eisenbud)?

This is testing my understanding of algebras, subalgebras, & polynomial rings. Page 31. Corollary 1.5. Let $k$ be a field, and let $S = k[x_1, \dots, x_r]$ be a polynomial graded by degree. Let $R$ be a $k$-subalgebra of $S$. If $R$ is a summand of $S$, in the sense that there is a map […]

Proof that $\mathbb{Z}$ has no zero divisors

Everyone knows the rules of zero divisors like $$\forall \alpha,\beta\in\mathbb{R}\;:\;\alpha\cdot\beta = 0\Rightarrow\alpha=0\vee \beta=0.$$ But how can I prove it for $\mathbb{Z}$? My first try was this one: For $\alpha\cdot \beta=0$ and $\alpha\neq 0$ let $$0=\alpha^{-1}\cdot 0=\alpha^{-1}\cdot (\alpha\cdot\beta)=(\alpha^{-1}\cdot \alpha)\cdot\beta = \beta = 0\Rightarrow \beta = 0;$$ and the same for $\beta\neq 0\Rightarrow \alpha=0$, however i realized […]

A particular isomorphism between Hom and first Ext.

Let $R$ commutative ring and $I$ an ideal of $R$. How do I prove that $\operatorname{Ext}^1_R(R/I,R/I)$ isomorphic to $\operatorname{Hom}_R(I/I^2,R/I)$ ? This question is an exercise of the course but has a chance of being false.

Subrings of polynomial rings over the complex plane

I have the following questions: (i) must every subring of the polynomial ring in two variables over the complex plane, containing the complex plane itself, be Noetherian? (ii) Are there Noetherian rings containing the complex field which are not f.g. as a ring over the complex field (maybe the field of rational functions over the […]

Deduction of usual Cayley-Hamilton Theorem from “Determinant Trick”

Here is a statement of a standard theorem in commutative algebra (see page 60 of this book): Theorem. (“Determinant Trick”) Suppose that $R$ is a commutative ring with $1$. Let $M$ be a finitely generated $R$-module, generated by $m_{1}, …, m_{n}\in M$. Let $\varphi: M\to M$ is an endomorphism, and assume that $\varphi(m_{i})=\sum_{j=1}^{n} a_{ij} m_{j}$ […]

When does the regularity of $A$ implies the regularity of $A$?

Let $A$ be a commutative noetherian ring (I do not mind to assume that $A$ is a UFD), and assume that $A$ is regular. Recall that a commutative noetherian ring is called regular if all its localizations at maximal ideals are regular local rings. Let $w$ be algebraic over $A$ (I do not mind to […]

If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$?

If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$? Background: I was trying to prove that if $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules then $m=n$. For this I need a step like $R{^m/M{^m}}$ isomorphic to […]

$A \subseteq B \subseteq C$ with $A \subseteq C$ separable and $pd_{B \otimes_A B}(B) = \infty$

Assume $A \subseteq B \subseteq C$ are commutative rings such that $C$ is separable over $A$, namely $C$ is a projective $C \otimes_A C$-module. Separability of $C$ over $A$ does not imply separability of $B$ over $A$. In other words, it can happen that $pd_{C \otimes_A C}(C)=0$ and $pd_{B \otimes_A B}(B) > 0$. Can one […]