Chủ đề này có 5 trả lời

[canh_dieu]

A friend of mine asked for an example of a non-commutative ring http://dientuvietnam...n/mimetex.cgi?R with a maximal ideal http://dientuvietnam...n/mimetex.cgi?M such that the quotient ring R/M is not a field. I found such an example but it is not a nice one since the maximal ideal of that ring is the trivial ideal . Any one has better ideas?

Notes.

1. For commutative rings such an example does not exist.

2. Here ideals in non-commutative rings are understood as two-sided ideals.

[vinhspiderman]

This is my own counter example (it contains a large class of examples because it is general!) :

Let R be a commutative ring which has more than 2 elements, M is its maximal ideal.
Now put V=RxR, we define on V + and * as follow :
(a,b)+(c,d)=(a+c,b+d)
(a,b)*(c,d)=(ad,bc)
It is easy to check that with + and *, V becomes a noncommutative ring which has a unit. As in the case of commutative ring, by using Zorn's lemma, we also prove that there exists at least one maximal ideal on V which contains a given ideal of V. Let M be a maximal ideal which contains the ideal generated by (0,1+1) (in some particular case, to show M clearly isn't an easy work!).
Now, from the facts that V is noncommutative and that (0,1+1) belongs to M, if V/M is commutative, we deduce that (1,-1) belongs to M, and induce (1,-1)+(0,1+1)=(1,1) belongs to M, a contradiction.
So V/M is noncommutative, and for that it is not a field.

Finally, it is easy to see that from this example, there are many nontrivial examples by choosing suitable particular commutative rings R.

[madness]

(a,b)*(c,d)=(ad,bc)

There is some problem here. If (c,d) is the unit of your ring, then (c,d)=(1,1). But (1,1)*(a,b)=(b,a) <> (a,b)?

Can the following be an example?

Let R be a ring and M=M_n( R ) be the ring of n_n matrices whose entries are elements in R.

Proposition 1: If I is an ideal of R, then M_n(I) is an ideal of R and M/M_n(I) ~ M_n(R/I). (~ : ring isomorphism)

Proposition 2: Every two sided ideal in M has the form M_n(I) for some ideal I of R.

Corollary: R is simple if and only if M is simple.

So you just choose an arbitrary simple ring R, then you get a non-commutative ring M, which has a maximal ideal which is the trivial ideal O, and M/0 is not a field.

If you don't want a trivial maximal ideal, then you can choose an arbitrary ring R (may be commutative) and an arbitrary maximal ideal I of R (there should be tons of examples for this purpose). You can claim that M_n(I) is a maximal ideal of M because R/I is simple and M/M_n(I) ~ M_n(R/I) is therefore a simple ring, thus has no ideals. But M_n(R/I) can be non-commutative, for example when R/I ~ Z/p, thus it cannot be a field.

Please correct me if I'm wrong

P/s: should this topic be put in the "Dai so and Ly thuyet so" box?

[vinhspiderman]

Sorry, I was wrong. Thank madness for point out my "stupid" mistake.

[canh_dieu]

I believe that madness's examples are correct. (My example was ) The two propositions look very cute . Since I have been working mainly with commutative rings, non-commuatative rings always scare me.

P/s. This topic should not be moved to anywhere else. We are trying to develop a box, at this time just for college students only, for those who prefer to post their questions, comments...in English.

[nemo]

I have a problem: Let r,s be elements of a semisimple ring R so that rs=1. Show that sr=1.

Can someone show me, any help would be greatly appreciated .