How are the maximal ideals of a semisimple ring related to the ideals of the center of the ring? Let $R$ be a semisimple ring, in the sense that the regular left module ${}_R R$ is semisimple. By the Wedderburn–Artin theorem I would expect there to be a bijection between the maximal (say left) ideals of $R$ and the maximal ideals of $\operatorname{Z}(R)$.
Is this correct? If so, is there a more direct way of obtaining this bijection without invoking Wedderburn–Artin?
 A: There is in general no bijection between the maximal left ideals of $R$ and the maximal ideals of $\mathrm{Z}(R)$.
Let us consider the special case of $R = \mathrm{M}(n, )$, were $$ is some field.
The center is $R$ is $$, which contains precisely one maximal ideal.
But the ring $R$ contains at least $n$ maximal left ideals: we have for every column index $j = 1, \dotsc, n$ a maximal left ideal given by all those matrices whose $j$-th column consists of zeroes.
For $n ≥ 2$ we have no bijection between maximal left ideals of $R$ and maximal ideals of the center of $R$.

We do, however, have a bijection between the maximal two-sided ideals of $R$ and the ideals of $\mathrm{Z}(R)$.
This can be seen from the Artin–Wedderburn theorem thanks to the following facts:

*

*The maximal two-sided ideal in a product of rings $R × S$ are precisely of the form $I × J$ where $I$ is a two-sided ideal of $R$ and $J$ is a two-sided ideal of $S$.

*For every skew field $D$ and size $n ≥ 1$, the only two-sided ideal in the matrix ring $\mathrm{Mat}(n, D)$ are $0$ and $\mathrm{Mat}(n, D)$.
In other words, $\mathrm{Mat}(n, D)$ is a simple ring.


I think this can also be established from the ground up as follows (sketch):

*

*The semisimple ring $R$ admits only finitely many simple modules $S_1, \dotsc, S_n$ up to isomorphism.

*For every $R$-module $M$, let $M_i$ be the $S_i$-isotypical component of $M$: the sum of all submodules of $M$ that are isomorphic to $S_i$.
We have $M = M_1 ⊕ \dotsb ⊕ M_n$, a decomposition of $R$-modules.
If $f \colon M \to N$ is a homomorphism of $R$-modules, then $f(M_i) ⊆ N_i$ for every $i = 1, \dotsc, n$.

*It follows that each $R_i$ is a two-sided ideal of $R$.
The decomposition $R = R_1 ⊕ \dotsb ⊕ R_n$ is therefore one of two-sided ideals.
It follows that each $R_i$ becomes a ring by restricting the multiplication of $R$ to $R_i$, and we then have $R ≅ R_1 × \dotsb × R_n$ as rings.
(Each ring $R_i$ is again semisimple, with $S_i$ being it’s only simple module up to isomorphism.)

*It can be shown that each $R_i$ is simple. (This may require a technical lemma like one of the Jacobi density theorems; I don’t quite remember off the top of my head.)

We have thus found a decomposition $R = R_1 × \dotsb × R_n$ into simple rings.
(This is basically a weak version of the Artin–Wedderburn theorem.)

*

*It now follows that the maximal two-sided ideals of $R$ are precisely $R_1 ⊕ \dotsb ⊕ \widehat{R_i} ⊕ \dotsb ⊕ R_n$ for $i = 1, \dotsc n$.

So what about the center?

*

*We have $R ≅ \mathrm{Z}(R_1) × \dotsb × \mathrm{Z}(R_n)$.


*Every ring $T$ can be seen as a $T$-$T$-bimodule.
The two-sided ideals of $T$ are precisely the sub-bimodules of $T$.
Therefore:

*

*$T$ is simple as a ring if and only if it is simple as a $T$-bimodule.

*If $M$ is a simple $T$-bimodule then $\mathrm{End}_{\text{$T$-bimod}}(M)$ is a skew field by (a version of) Schur’s lemma.

*The endomorphism ring of $T$ as a $T$-bimodule is precisely $\mathrm{Z}(T)$.

We hence see that for a simple ring $T$, its center is a skew field.


*We see that $\mathrm{Z}(R_i)$ is a skew field for every $i = 1, \dotsc, n$.


*The maximal ideals of $\mathrm{Z}(R)$ are therefore precisely $\mathrm{Z}(R_1) × \dotsb × \widehat{\mathrm{Z}(R_i)} × \dotsb × \mathrm{Z}(R_n)$ for $i = 1, \dotsc, n$.
