# Tag Info

7

Here's a counterexample. Let $R=\mathbb{Z}[2^{1/3},2^{1/9},2^{1/27},\dots]\subset\mathbb{R}$. Then $R$ is not Noetherian, since the ideal $(2^{1/3},2^{1/9},2^{1/27},\dots)$ is not finitely generated. But the only ring-homomorphism $f:R\to R$ is the identity. Indeed, any such $f$ must be the identity on $\mathbb{Z}$, and then $f\left(2^{1/3^n}\right)$ ...

5

A commutative artinian ring is a finite product of local artinian rings. If the only units in $R$ are $1$ and $-1$, then the same is true of any direct factor of $R$. If $S$ is a local artinian ring with maximal ideal $m$, then every element of $S\setminus m$ is a unit. If the only units are $1$ and $-1$, this means that either $m$ has $1$ element and $S$ ...

2

Take $R = \mathbb{F}_2[X]/(X^n)$. Then $R$ is finite and has exactly $n+1$ ideals. Indeed, ideals of $R$ are in canonical bijection with ideals of $\mathbb{F}_2[X]$ containing $X^n$, ie with polynomials dividing $X^n$ : these are the $X^k$ for $0\leqslant k\leqslant n$.

2

Yes, $d(n)$ can attain any positive integer for some $n$, consider $n=2^k$. Hence all ring numbers are of the form $d(n)$.

2

Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$b_1-b_2=j_2-j_1\in B\cap I,$$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto. As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, ...

2

All non-zero ideals of $R=\mathbb Z$ are isomorphic. Are the corresponding quotients isomorphic?

2

If your nil ideals are two-sided proceed like this: Let $I,J$ be two-sided nil ideals of a Ring $R$. We want to show that $I+J$ is also nil. To do this consider $R/J$ and $(I+J)/J$. You can see rather easily that $(I+J)/J$ is a two-sided nil ideal of $R/J$: Let $[x] \in (I+J)/J$, then there exists a representative $a$ of $[x]$ that lies in $I$. Since then ...

1

1) $S$ is also a left ideal, in fact, if $r\in R$ and $I$ is a right ideal of $R$ so is $rI$ (Evident!). Let $x\in I$, we first observe that $(xr)^k=0$ for some integer $k$. Therefore, we have $(rx)^{k+1}=0$, showing that $rI$ is a nil right ideal of $R$. Now, if $y\in S$ we can write $y=y_1+...y_n$, where each $y_j$ belongs to some nil right ideal $I_j$ ...

1

To show that there are infinitely many positive integers which are not principal it suffices to exhibit infinitely many positive integers $n$ such that there exists a non-principal ring with $n$ ideals. This is straightforward and can be done in many ways; here's one. Let $R = \mathbb{F}_q[x, y]/(x^2, xy, y^2)$. This ring is not principal because $m = (x, ... 1 It's obvious that$(a)+(b)\subseteq R$. The other inclusion in the point of the question, and you (almost) proved it! From$as+bt=1$we get$asr+btr=r$for any$r\in R$, so$r\in (a)+(b)$. 1 Well, it seems you have all needed to conclude. Let's start with the end of the proof: one knows that$r=\min\{i:H_i({\bf x,y};M)\ne0\}$. Set$J=({\bf x,y})$. Then$J$contains a maximal$M$-sequence, namely${\bf y}$. If$JM\ne M$then one can apply the Corollary 17.12 and get$s=\min\{i:H_i({\bf x,y};M)\ne0\}$, so$s=r$. The only thing to show now is ... 1 Here is a link to a textbook that adopts the convention$\mathfrak{p}^0 = R$explicitly (see the bottom line). 1 Since$M=N+IN'\subseteq N+IM$you get$M=N+IM$. Can you continue from here? Since$M=N+IN'\subseteq N+N'$you get$M=N+N'$. Can you see now why$M/N$is finitely generated? 1 Here is an example for question 2: Let$J$be an ideal, and$M=A/J$. The$M\otimes I\simeq I/IJ$.$IM=I\cdot A/J=(I+J)/J\simeq I/(I\cap J)$Now usually,$IJ\neq I\cap J$, unless$I$and$J$are coprime, i.e.$I+J=A$. In particular, if$J=I$,$A/I\otimes I\simeq I/I^2$, while$I\cdot A/I=0$. 1 For question 1, let$A = k[x, y]/(x^2, xy, y^2)$and$I = (x, y)$. The$A$-module endomorphisms$\text{End}_A(I)$can be identified with$2 \times 2$matrices, and of those, only the scalar matrices show up as multiplication by an element of$A$. For question 2, this is true for every ideal$I$if and only if$M$is flat. This comes from thinking of$IM\$ as ...

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