8 votes

Noetherian rings definition

I think your confusion stems from the following fact: A maximal element isn't necessarily a greatest element. Maximal only means nothing is above it. It does not mean it's above everything else. You ...
Arthur's user avatar
  • 200k
6 votes

Is $\mathbb{Q}/\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Q} = 0$?

No, this is incorrect. All the relations you wrote down are true, but there are more relations, because $$ \frac{a}{b} \otimes \frac{c}{d} = \frac{a}{b} \otimes \frac{bc}{bd} = \frac{ba}{b} \otimes \...
hunter's user avatar
  • 30.4k
4 votes

Noetherian rings definition

To prove $(2)\implies (1)$, let $\mathcal{F}$ be a nonempty set of ideals. Assume it has no maximal element. Take any $I_1\in\mathcal{F}$. It is not maximal, and so there is some $I_2\in\mathcal{F}$ ...
Mark's user avatar
  • 40.4k
2 votes

When will $\mathbb{Z}$ be a projective $\mathbb{Z}G$-module?

Write $x=\mu(1)=\sum x_gg$. Since $(g-1)x=0$, we must have $x_g=x_1$ for all $g$, i.e. $x=x'(\sum_{g\in G} g)$ for some $x'\in \mathbb Z$. This implies that $x'|G|=1$, so this can only be the case ...
Ben Webster's user avatar
  • 1,041
2 votes

A book to review abstract algebra?

Two books I can vouch for are Abstract Algebra by Pinter and Abstract Algebra: Theory and Applications by Thomas Judson. The former is what I would consider to be the "definitive" self-study ...
Grey's user avatar
  • 185
1 vote

I have a question of Prop.9.2 of Atiyah's commutative algebra.

You want to show that $M/M^2$ is at most a one-dimensional space over the field $k=A/M$. Since we are assuming (in that proposition) that $M$ is principal ideal, we can find $g\in M$ so that $M = Ag$. ...
Nicolas Bourbaki's user avatar
1 vote

Non radical ideal with maximal radical

There are plenty of easily accessible examples. For instance, $(X^2)\subseteq \Bbb R[X]$, or $(125)\subseteq \Bbb Z$. Often enough, you can construct an example by taking a ring $R$ and a maximal ...
1 vote

What is the difference between free product and direct sum?

Let's say that we have two groups: $G=H=\mathbb{Z}/2\mathbb{Z}$. So these are finite groups of order $2$. Their direct sum $G\oplus H$ is their Cartesian product with pointwise group operation. In ...
freakish's user avatar
  • 43.2k
1 vote

Modules with no regular elements in his annihilator

Converting comment-solution by user26857 into a real solution: The equality is wrong. Take a local ring $(𝑅,𝑚)$ which is not Cohen-Macaulay and is of dimension 1, and let $𝑀=𝑅/𝑚$ as a ...

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