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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.

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Integral closure of $k[x^4,x^3y,xy^3,y^4]$

Let $R=k[x^4,x^3y,xy^3,y^4]$. We can see $R$ is not integrally closed (since $x^2y^2 \in Q(R)$ is integral over $R$ but $x^2y^2 \notin R$ ). Therefore $k[x^4,x^3y,x^2y^2,xy^3,y^4]\subseteq\overline{R}$ (where $\overline{R}$ is integral closure). How…
user177523
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Height of $I$ = Height of $I K[X]$

In the proof of Theorem 23 in Matsumura's Commutative Algebra on page 85, he wrote Since $A$ is a subring of $B = A[X]/I$, we have $A \cap I = (0)$. Therefore, if $K$ denotes the quotient field of $A$ then $\text{ht}(I) = \text{ht}(IK[X]) \leq \dim…
An Hoa
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Integrally Closed implies reduced

Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?
messi
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When does $v \wedge w = w \wedge v$ imply $v \wedge w=0$?

Let $A$ be a noetherian ring such that $2$ is not a zero divisor of $A$. Let $M$ be a submodule of the free module $F=A^r$. Clearly $2$ is not a zero divisor of $M$, i.e. $2x=0$ implies $x=0$ for all $x \in M$. But how is the situation for the…
Hans
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For a reduced ring $A$, must $A[x]\setminus A$ be multiplicatively closed?

Suppose $A$ is a reduced commutative ring. Is $A[x]\setminus A$ is multiplicatively closed?
messi
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Polynomial algebra

Let $k$ be a field and $k \subset A \subseteq k[X]$ be a $k$-subalgebra of $k[X]$. Prove that $\dim(A)=1$ (Krull dim) and that $A$ is a finitely-generated $k$-algebra. My initial thought: Consider the short exact sequence $0 \rightarrow A…
Adrian Manea
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Quotient field of a domain

Let $A$ be a commutative domain and $K=Quot(A)$, its field of fractions (quotient field). Prove that $K$ is a f.g. $A$-module if and only if $A=K$.
Adrian Manea
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Independence of Exercises in Atiyah and MacDonald

I'm reading Atiyah and MacDonald, and I would like to get through the whole book. There are some exercises in Chapter 2 in Atiyah and MacDonald about absolutely flat rings. However, it is in the section that uses the Tor functor, which I don't know…
Chris Z
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An example of a primeless (i.e. module without prime submodule) and projective module

Please, give an example of a module $M$ such that $M$ is primeless (i.e. without prime submodule) and projective. Thanks for your attention.
m. sam.
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There is no unique continuous homomorphism

Are there any hints to solve the exercise 1.3.9(c) in Qing Liu's book Algebraic Geometry and Arithmetic Curves? Let $n\geq 2$ be an integer, and $D=\mathbb Z[1/n]$. Let $A$ be a complete commutative ring with unit for the $I$-adic topology, where…
selfstudying
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Well-ordering of Graded reverse lexicographic order

I am stuck with yet another question to "show that grevlex is a monomial order" I have done everything apart from showing it is "well-ordered" I've looked at a proof for "grlex" for reference but it didn't help me much; it uses the fact that "lex"…
Kydo
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Modulus of n-tuples?

I've tried Googling first but I didn't really find a clear answer. I am looking at the definition of "Graded Lex Order" where it says Let $\alpha$,$\beta \in \mathbb{Z}^n$. We say $\alpha >_{grlex} \beta$ if $|\alpha|=…
Kydo
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Prime Ideals in a Localised Ring

Let $R$ be a commutative ring, and let $S \subseteq R$ be a multiplicatively closed subset (not containing $0$). Then we construct the localised ring $R [ S^{-1} ]$. I understand that prime ideals in $R [ S^{-1} ]$ correspond to prime ideals in $R$…
Paul Slevin
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Going down fails when $R$ is not integrally closed.

In this post there is a counterexample to the going down theorem. I am pretty sure that the reason why it fails is because $R$ is not integrally closed in $A$, but I don't have any nice argument to show that this is true. I would appreciate any…
math635
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Example where $\operatorname{grade}(I,M)>\operatorname{height} I$

Let $I$ be an ideal of a noetherian ring $R$ and let $M$ be a finite $R$-module. We need to show if $I$ is generated by $n$ elements, then $\operatorname{grade}(I,M)\le n$. Could any one give an example where…
Myshkin
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