# Tag Info

4

The most natural family of rings which comes to mind as having every commutative ring as a quotient is $\mathbb{Z}[\{t_i \mid i \in I\}]$, i.e., a polynomial ring in an arbitrary set of indeterminates over $\mathbb{Z}$. These rings are all UFDs -- see e.g. Corollary 15.27 of these notes -- hence integrally closed ("normal").

3

I believe this is the idea: Look in the local ring at the origin. Mod out by $x_2-x_1$. The quotient ring is now the localization of $k[x_2,x_3,x_4]/(x_2^2,x_2x_3,x_3x_4,x_2x_4)$, and everybody in the maximal ideal is now a zero-divisor. So the depth at the origin is only $1$.

3

Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that $M$ is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3.15 as well. Take $A = \mathbb{Z}$, and let $M$ be the direct sum of $\mathbb{Z}/k\mathbb{Z}$ as $k$ ranges through $\mathbb{N}$. This is not finitely ...

2

Hint: try $R=\Bbb R[[x]]$, and $S=\{1\}$ and you will find your answer. Complements of prime ideals necessarily have the property of being "saturated" in the sense that divisors of elements in the complement are in the complement.

2

(i) If $a\neq0$ the prime ideals you want are the maximal ideals $(x-a,y-b)\; (b\in \mathbb C)$ and $(x,y-b)\; (b\in \mathbb C)$ about which you already know PLUS the two prime but not maximal ideals $(x)$ and $(x-a)$. (ii) If $a=0$ the prime ideals you want are the maximal ideals $(x,y-b)\; (b\in \mathbb C)$ PLUS the prime ideal $(x)$ Thinking ...

2

Set $R = \mathbb{C}[x_1, \ldots, x_n]$. If $I \subset R$ is an ideal such that $\dim_{\mathbb{C}}(R/I) < \infty$ then $R/I$ is Artinian, and consider a descending sequence of ideals $\mathfrak{m}_1 \supset \mathfrak{m}_1 \cap \mathfrak{m}_2 \supset \cdots$ where $\mathfrak{m_i} \subset R$ is maximal. If $I \subset R$ is an ideal that is contained in ...

2

In the article they borrow the term from, the authors use $t$-disjoint to mean the edges are at distance at least $t$. Two edges in this particular graph $G$ that induce a subgraph of two disjoint edges are actually at distance at least $3$ (as opposed to $2$, as you might think), because the graph is bipartite.

2

Take $A = M = \mathbb{Z}_{(p)}$, localized at a prime $p$, and $N = \mathbb{Q}$. Take $f: M \to N$ to be the inclusion $\mathbb{Z}_{(p)} \hookrightarrow \mathbb{Q}$. Take the ideal $\mathfrak{a}$ to be the maximal ideal of $A$. In fact take any local integral domain $A$ and take $N$ to be its field of fractions. Incidentally (amusing) together with ...

1

Not every ideal of $R[x]$ is of the form $I[x]$ for some ideal $I\subset R$, so your proof is lacking. Moreover, your argument doesn't go through for the ideal $(X)\subset\Bbb{Z}[X]$, for example.

1

This is because a finitely generated projective module is finitely presented, so that for any prime ideal $\mathfrak p\in\operatorname{Spec} A$, $$\bigl(\operatorname{Hom}_A(P,A)\bigl)_{\mathfrak p}\simeq\operatorname{Hom}_{A_\mathfrak p}(P_{\mathfrak p},A_{\mathfrak p})$$ and also $\;(E\otimes_AF)_{\mathfrak p}\simeq E_{\mathfrak p}\otimes_{A_\mathfrak ... 1 Let's observe that if$a \neq 0$the ideals$(x)$and$(x-a)$are coprime, by CRT you obtain: $$\frac{\mathbb{C}[x,y]}{x(x-a)} \simeq \frac{\mathbb{C}[x,y]}{(x)} \oplus \frac{\mathbb{C}[x,y]}{(x-a)} \simeq \mathbb{C}[y]\oplus \mathbb{C}[y]$$ Now is easy to look at the prime ideals of$\mathbb{C}[x]\oplus \mathbb{C}[x]\$ and I think you can easely ...

1

There is a counter-example due to Ogoma in "Non-catenary pseudo-geometric normal rings", which seems to be a modification of one of Nagata's examples of a non-catenary ring. Heitmann has a simpler description of the construction in "A non-catenary, normal, local domain", as does Lech in "Yet another proof of a result of Ogoma".

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