Let R be a ring. Prove that there is a one-to-one correspondence between the ideals of R and the ideals of .

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- October 16th 2011, 06:30 PMkierkegaardIdeals of a ring
Let R be a ring. Prove that there is a one-to-one correspondence between the ideals of R and the ideals of .

- October 16th 2011, 07:38 PMDrexel28Re: Ideals of a ring
- October 16th 2011, 11:36 PMDevenoRe: Ideals of a ring
you know, i can see at once that is an ideal of whenever I is an ideal of R, what's less obvious to me is that every ideal of the matrix ring arises in this way.

the only thing i could think of is defining the ideal I from an ideal U in the matrix ring by:

(i was sooo tempted to use the identity matrix, but maybe R doesn't have unity).

but, i dunno if this is correct. inform me, alex. - October 17th 2011, 12:16 AMDrexel28Re: Ideals of a ring
I'm actually pretty sure you need to assume that is unital. The idea is basically to take some ideal, call it in , and let be the set of all elements of which appear as some entry in a matrix in . To prove this is an ideal you merely need to prove that for any one has that whose general is is in (this follows immediately by just applying on the left and right, NOTE this is where it's important we're dealing with two-sided ideals, certain matrices). This makes it obvious because given and one has that and so by assumption that is an ideal one has that , and are all in which tells us that and so really is an ideal. So, now we need to prove that . The one containment is by definition and the other follows for if then we know that for each and so by prior comment and so .

- October 17th 2011, 01:00 AMDevenoRe: Ideals of a ring
my understanding is, the matrix ring is defined even if R is not unital (think of all nxn matrices with even integer entries).

now, i understand "your" ring ideal is "bigger" than mine, because you are letting your ideal consist of elements that occur as any arbitrary entry.

(i thought of that, too, but didn't know how to specify it less vaguely). and i get at once that your definiton does yield an ideal.

the question for me, then becomes: is your ideal "strictly" bigger than mine? i'm not sure, in the general case, in the unital one, i'm pretty sure it isn't.

here's my reasoning: let's say that a is in A (for alex's ideal). so a is, oh, say the i,j-th entry in some matrix T. now we have the whole of the matrix ring

to use via multiplication to apply to T, and we'll still be in our matrix ring ideal (call it U, so our matrix T is in U).

if R is unital, then if i multiply T on the left with Ei,j (1 in the j-th position in the i-th row, 0's elsewhere),

i'll knock out everything but the i-th row. if i multiply this new matrix on the right by Ei,j, i'll obtain a(Ei,j).

now by suitably choosing Ek,l, i can also move (via left and right multiplication) the ai,j entry of T to any position i want.

so i can create a matrix a(Ei,i) for every value of i, and all of these are still in U. add them up, and you get a diagonal matrix.

but that means your ideal is contained in mine, so we actually have the same one.

if R is not unital, though, i can't do this....do we still have the same ideal? - October 17th 2011, 08:14 PMDrexel28Re: Ideals of a ring