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I don't understand this proof.

Before I read this,I think if R is not commutative,semilocal ring still has finitely many maximal left ideals.Because the following:

1.left Artin ring has finitely many maximal ideals.

2.There has a one to one correspndence between the maximal ideals of R which contains I and maximal ideals of R/I,where I is an ideal of R.

where am I wrong?I still don't understand the example given in the Remark.what is essential difference if we don't consider commutative ring.Thank you in advance!

Jian
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1 Answers1

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You're wrong here:

left Artin ring has finitely many maximal [left] ideals.

I suppose you meant "maximal left ideals" since that is what the text is talking about.

For example, $M_2(\mathbb R)$ has infinitely many maximal left ideals. Consider the family of matrices of the form $A_\lambda = \begin{bmatrix}1&\lambda\\0&0\end{bmatrix}$ for $\lambda\in\mathbb R\setminus \{0\}$. The set of principal left ideals, each one generated by one of these matrices, forms a set of pairwise distinct maximal left ideals.

Obviously you can replace $\mathbb R$ with any infinite field and the argument still works.

rschwieb
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  • thank you,this helps me a lot.but Istill has some difficults to understand this:1,commutative Artin ring has finitely many maximal ideal by using minimal condtion of all finite interaction of maximal ideals.Why can't we use this for left Artin ring? Is minimal condition not true for left Artin ring? 2.Left Artin may have infinite distinct maximal ideal as you give.Is Left Artin ring has finitely nonisomorphic maximal ideal? – Jian Oct 01 '17 at 12:15
  • @Sky 1,commutative Artin ring has finitely many maximal ideal by using minimal condtion of all finite interaction of maximal ideals. I don't really see why that says there are finitely many maximal ideals. For me it's much easier to say a commutative Artinian ring is a finite product of $n$ local rings, and then obviously there are only $n$ maximal ideals. In fact, every finite intersection (of more than one) of the left ideals I gave is zero, and you can see that doesn't imply there are finitely many of them. – rschwieb Oct 01 '17 at 14:20
  • @Sky 2.[...] Left Artin ring has finitely nonisomorphic maximal ideal? Not necessarily. In the example I gave, all maximal ideals are mutually isomorphic. You can make ones with nonisomorphic maximal left ideals, though. I'm not sure if it is possible to have infinitely many isoclasses of maximal left ideal in a left Artinian ring. – rschwieb Oct 01 '17 at 14:22
  • yeah,you are right.this may also give an example:R is not commutative,I+J=R,but IJ is not equal to $I\bigcap J$.From this,maybe commutative ring has biggest difference with non-commutative ring. – Jian Oct 01 '17 at 14:34
  • @rschwieb (last sentence): I don't think that is possible, because the maximal left ideals are in bijection with those in $R/J(R)$, which is semisimple for left Artinian $R$, and in semisimple rings there are only finitely many isoclasses of maximal left (or right) ideals. – Torsten Schoeneberg Oct 02 '17 at 22:48
  • @TorstenSchoeneberg Oh of course, I forgot the original context was confined to semilocal rings and I was just considering general rings. Yes, that's completely right. – rschwieb Oct 02 '17 at 23:41
  • @TorstenSchoeneberg today,I think this question again.I don't know how to prove there are finitely many isomorphic class of maximal ideals in Artin ring.Semisimple ring has finitely many maximal ideals.For artin ring R,we can get $m_i/J$ has finitely many isomorphic class where {$m_i$}are all the maximal ideals of R.How to prove?Thank you in advance. – Jian Nov 28 '17 at 00:28
  • @Sky "Semisimple ring has finitely many maximal ideals" for a commutative semisimple ring, yes but it is false for noncommutative semisimple rings. – rschwieb Nov 28 '17 at 01:09
  • @Sky For any ring $R$, $R$ and $R/J(R)$ share the same simple right modules, where $J(R)$ is the Jacobson radical. If $R$ is Artinian, $R/J(R)$ is semisimple, and hence has only finitely many isotypes of simple right modules. Therefore $R$ only has finitely many isotypes of simple right modules. – rschwieb Nov 28 '17 at 01:10
  • @rschwieb yeah，like matrix ring.I forget to say isomorphic class. – Jian Nov 28 '17 at 01:30
  • @rschwieb artin ring has finitely many isotypes of simple modules.but this can't show that artin ring has finitely many isotypes of maximal ideals. – Jian Nov 28 '17 at 01:31
  • @Sky you don’t need isotypes of maximal ideals, although they correspond 1-1 with isotypes of simple modules. “But this can’t show artin ring has finitely many isotypes of [simple modules]” yes, it does, for the reason I gave before. Do you have a question about the reason? – rschwieb Nov 28 '17 at 02:50
  • @rschwieb you gave the reason it has finitely many isotypes of simple modules.But I don't know why the isotypes of maximal ideals is 1-1 correspondence with the isotypes of simple modules. – Jian Nov 28 '17 at 04:41
  • @Sky OK, I see, I can explain further on that. The strategy is mentioned several places on the site. See [this problme(https://math.stackexchange.com/q/325825/29335) for some explanations. Because $J(R)$ is contained in every maximal right ideal, it annihilates every simple right $R$ module, so this solution explains why the simple $R$ modules and the simple $R/J(R)$ modules are the same. – rschwieb Nov 28 '17 at 12:09
  • @rschwieb Maybe you don't understand where I have difficulty.my question:for different maximal ideal $m_1$ and $m_2$,$if R/m_1\cong m_2$,we can't get $m_1\cong m_2$.So how can we show that the isotypes of maximal ideal is 1-1 correspondence with the isotypes of simple modules? – Jian Nov 28 '17 at 12:51
  • @Sky The post I cited in the last comment tells you exactly that. It shows you that every simple $R$ module has a $R/J(R)$ module structure and vice versa. Consequently, an abelian group $S$ is a simple $R$ module iff it is a simple $R/J(R)$ module. The sets of abelian groups which are simple modules for each ring are equal. – rschwieb Nov 28 '17 at 13:34
  • @rschwieb I always know that S is simple R module iff it is simple R/J(R) module.But I can't construct the connection between isomorphic class of maximal ideal and isomorphic class of simple R modules – Jian Nov 28 '17 at 13:54
  • @rschwieb such as for two isomorphic modules $m_1$ and $m_2$,I think we even don't know whether $R/m_1$ is isomorphic to $R/m_2$ or not. – Jian Nov 28 '17 at 14:04
  • @sky But that is unnecessary. If $S\cong S'$ is an isomorphism of simple $R$ modules then you know that $S\cong S'$ as simple $R/J(R)$ modules. The exact same isomorphism works for both module actions. Since the simple modules correspond, the isoclasses correspond. If you try to write it all down I think you'll see more clearly. – rschwieb Nov 28 '17 at 14:17
  • @rschwieb but if S is not simple R-module,S may not be R/J(R) module.for maximal ideal of R.it is natural that m/J(R) is R/J(R) module.maybe I understand wrong for what you have said. – Jian Nov 28 '17 at 14:33
  • @Sky Everything you said in your last comment seems correct, I just don't understand why you think it is relevant. We're only talking about simple modules, so it seems irrelevant to talk about ones that aren't simple. – rschwieb Nov 28 '17 at 14:38
  • @rschwieb my original question:Why left atrin ring has finitely non-isomorphic maximal left ideals? Please forgive my poor English expression. – Jian Nov 28 '17 at 14:43
  • @Sky Isomorphism of maximal left ideals has nothing to do with isomorphism of simple modules. I don't have any comments about it because it isn't related to what we're doing. – rschwieb Nov 28 '17 at 14:48
  • @rschwieb oh,you can see torsten-schoeneberg's comment.and you also say completely correct.maybe I misunderstand. – Jian Nov 28 '17 at 14:51
  • Thank you for your commutation. – Jian Nov 28 '17 at 14:52
  • @Sky The maximal right ideals themselves are in bijection because of ideal correspondence. I can't say for sure if their isoclasses coincide too, but perhaps. I think perhaps it was a mild inaccuracy in what he was saying. Nevertheless, he got the point across to me that there had to be the same number of simple module isoclasses. – rschwieb Nov 28 '17 at 15:00
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    I hope I do not add to misunderstandings here. Let me clarify my last comment in three steps: 1) $m \mapsto m+J(R)$ induces a bijection between the maximal left ideals of $R$ and the maximal left ideals of $R/J(R)$ (not only isoclasses). That is essentially because every maximal left ideal of $R$, by definition, contains $J(R)$. -- Because of this, we can wlog assume that $R$ is (artinian) semisimple. 2) We all agree, now, that over an (artinian) semisimple ring, there are only finitely many isoclasses of simple modules. [to be continued ...] – Torsten Schoeneberg Nov 28 '17 at 21:13
  • By definition of (artinian) semisimple, every left module over $R$ is semisimple, and every f.g. left module is Artinian semisimple. In particular, every left ideal of $R$ -- maximal or not -- is Artinian semisimple, hence isomorphic to a finite direct sum of simple modules. With Artin-Wedderburn and for dimension reasons, the number of summands here is actually bounded from above. This combined with 2) implies that there only finitely many isoclasses of left ideals in $R$. [to be continued ...]
    • – Torsten Schoeneberg Nov 28 '17 at 21:27
  • Concretely, every left ideal of $R\simeq M_n(D)$ is isomorphic to the direct sum of $p$ copies of $V$, where $0\le p \le n$ and $V$ is "the" simple module $D^n$. So if $R\simeq\prod_{i=1}^k M_{n_i}(D_i)$, and $V_i:=D_i^{n_i}$, every left ideal of $R$ is isomorphic to some "sub-sum" $\bigoplus_{i=1}^k V_i^{p_i}$ with $0\le p_i\le n_i$. The maximal left ideals in particular are those where all $p_i=n_i$ except for one index $i_0$ with $p_{i_0}=n_{i_0}−1$. -- The punchline in 3 is that due to semisimplicity, quotients "are" submodules and submodules "are" quotients, up to isomorphism. – Torsten Schoeneberg Nov 29 '17 at 18:37