Let R be a ring. Prove that there is a one-to-one correspondence between the ideals of R and the ideals of $\displaystyle M_{n}(R)$.
you know, i can see at once that $\displaystyle \text{Mat}_n(I)$ is an ideal of $\displaystyle \text{Mat}_n(R)$ 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:
$\displaystyle I = \{a \in R\ |\ \text{diag}\{a,a,\dots,a\} \in U\}$
(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.
I'm actually pretty sure you need to assume that $\displaystyle R$ is unital. The idea is basically to take some ideal, call it $\displaystyle \mathfrak{b}$ in $\displaystyle \text{Mat}_n(R)$, and let $\displaystyle \mathfrak{a}$ be the set of all elements of $\displaystyle R$ which appear as some entry in a matrix in $\displaystyle \mathfrak{b}$. To prove this is an ideal you merely need to prove that for any $\displaystyle x\in\mathfrak{a}$ one has that $\displaystyle xO_{i,j}$ whose general $\displaystyle (\ell,k)^{\th}$ is $\displaystyle x\delta_{i,\ell}\delta_{j,k}$ is in $\displaystyle \mathfrak{b}$ (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 $\displaystyle x,y\in\mathfrak{a}$ and $\displaystyle r\in R$ one has that $\displaystyle xO_{i,j},yO_{i,j}\in\mathfrak{b}$ and so by assumption that $\displaystyle \mathfrak{b}$ is an ideal one has that $\displaystyle rO_{1,1}xO_{1,1}=(rx)O_{1,1}$, $\displaystyle xO_{1,1}rO_{1,1}=(xr)O_{1,1}$ and $\displaystyle xO_{1,1}-yO_{1,1}=(x-y)O_{1,1}$ are all in $\displaystyle \mathfrak{b}$ which tells us that $\displaystyle rx,xr,x-y\in\mathfrak{a}$ and so $\displaystyle \mathfrak{a}$ really is an ideal. So, now we need to prove that $\displaystyle \mathfrak{b}=\text{Mat}_n(\mathfrak{a})$. The one containment is by definition and the other follows for if $\displaystyle M=(m_{\ell,k})\in\text{Mat}_n(\mathfrak{a})$ then we know that $\displaystyle m_{\ell,k}\in\mathfrak{a}$ for each $\displaystyle m_{\ell,k}$ and so by prior comment $\displaystyle m_{\ell,k}O_{\ell,k}\in\mathfrak{b}$ and so $\displaystyle \displaystyle \sum_{\ell,k}m_{\ell,k}O_{\ell,k}=M\in\mathfrak{b}$.
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?