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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!

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Jian
    Oct 1, 2017 at 12:15
  • $\begingroup$ @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. $\endgroup$
    – rschwieb
    Oct 1, 2017 at 14:20
  • $\begingroup$ @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. $\endgroup$
    – rschwieb
    Oct 1, 2017 at 14:22
  • $\begingroup$ 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. $\endgroup$
    – Jian
    Oct 1, 2017 at 14:34
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    $\begingroup$ 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 ...] $\endgroup$ Nov 28, 2017 at 21:13

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