As mentioned in the previous article, we will now describe the “bad elements” in a ring R which stops it from being semisimple. Consider the following ring: $R = \left \{ \begin{pmatrix} a & b \\ 0 & d\end{pmatrix} : a,b,d\in \mathbf{R}\right\}$

Since R is finite-dimensional over the reals, it is both artinian and noetherian. However, R is not semisimple since the module MR2 (whose elements are column vectors) with R acting by left-multiplication is not semisimple. Indeed, the R-space spanned by (1, 0) is a submodule which is not a direct summand.

Definition. A maximal submodule of M is an $N\subsetneq M$ such that if N’ satisfies N ⊆ N’ ⊆ M, then N’=N or N’=M.

For example, the zero module has no maximal submodules. For any prime p, the Z-submodule pZ ⊂ Z is maximal. Also, if $M = \{\frac a {2^m} : a, m\in \mathbf{Z}\}$, then the Z-module M/Z has no maximal submodule.

Exercise: prove the last sentence in the previous paragraph.

Now we shall define what we mean by “bad elements”.

Definition. The radical of M, written as rad(M), is the intersection of all maximal submodules of M. When M=R, this is also called the Jacobson radical and denoted J(R). Let us consider the case of modules first. We have the following basic result.

Lemma. Let M, N be modules over a fixed ring R.

• If p : M → N is surjective, then p(rad(M)) ⊆ rad(N).

Proof

• If x∈rad(M) and P is a maximal submodule of N, then M ∩ P ⊆ M is either maximal or equal to M (since we get an injection M/(M ∩ P) → N/P into a simple module). In either case, x lies in M ∩ P ⊆ P. Hence x∈rad(N).
• Let x∈rad(M) and P be a maximal submodule of N. Then p-1(P) ⊂ M is maximal too since M/p-1(P) is isomorphic to N/P. Thus xp-1(P) and p(x)∈P.
• ⊇ follows from the first part and M⊕0, 0⊕N ⊆ MN. And ⊆ follows from the second part, applied to projections MN → M and MN → N. ♦

The next result relates the radical to semisimplicity.

Proposition. If M is non-zero and semisimple, then rad(M) = 0. In particular, the Jacobson radical of a semisimple ring is 0.

Proof

If M is semisimple, write M as a direct sum $\oplus_i M_i$ of simple submodules $M_i \subseteq M.$ For each index i, let $N_i$ be the submodule of M obtained by summing all $\{M_j\}_{j\ne i}.$ Then each $N_i$ is a maximal submodule of M and clearly $\cap_i N_i =0.$ ♦

Corollary. Since we know that a semisimple ring is artinian and noetherian, we conclude that a semisimple ring R is artinian and its Jacobson radical is zero.

Now, we will prove the converse. Here is an alternative way of describing the Jacobson radical of R.

Lemma. J(R) is the set of all x∈R for which xM = 0 for any simple left module M.

Proof

Suppose xM = 0 for any simple M. If I ⊂ R is a maximal left ideal, then R/I is simple, so x(R/I) = 0 and taking the image of 1 in R/I, we get xI. So x is contained in every maximal left ideal of R.

Conversely, suppose x is in J(R). Any simple left module M is a quotient R/I, where I is a maximal left ideal. [ Warning: it is tempting to conclude xI, and so x(R/I) = 0, but the 2nd conclusion is not obvious due to the left/right subtlety! ]

Pick any yR. Since R/I is simple, the image of y in R/I is 0 or generates the whole module. In the former case, yI so xyI. Otherwise the map fR → R/I mapping 1 to y+I is surjective, so it induces an isomorphism gR/J → R/I. Then J is maximal in R, so xJ, and we get 0 = g(x+J) = g(x(1+J)) = xy+I, and so xyI. ♦

Note

Since every semisimple module is a sum of simple submodules, the lemma also shows that xR iff xM = 0 for any semisimple left module M.

Corollary. J(R) is a two-sided ideal of R.

Proof

We already know J(R) is a left module, being an intersection of left modules. Let xJ(R), yR; we need to show xyJ(R). But for any simple module M, we have yM ⊆ M so xyM ⊆ xM = 0 since xJ(R). Hence we also have xyJ(R). ♦

Exercise.

In the above proof to the corollary, is it correct to say: yM ⊆ M must be 0 or whole of M since M is a simple module? [ Answer: no, because yM may not be a left submodule of M. ]

Example

Consider the ring $R = \left\{ \begin{pmatrix}a & b\\ 0 & d\end{pmatrix} : a, b, d\in\mathbf{R}\right\} = \overbrace{\left\{ \begin{pmatrix} a & 0\\ 0 & 0\end{pmatrix} : a\in\mathbf{R}\right\}}^{I} \oplus \overbrace{\left\{ \begin{pmatrix} 0 & b\\ 0 & d\end{pmatrix} : b,d\in\mathbf{R} \right\}}^{J}$

written as a direct sum of two left ideals. Clearly I is simple since it has dimension 1 over R. On the other hand, J has a simple submodule $J' = \left\{ \begin{pmatrix} 0 & b\\ 0 & 0\end{pmatrix} \right\}$ which is isomorphic to I, and the quotient J/J’ is another simple module. Hence any simple module of R is isomorphic to I or J/J’.

[ To see why the last statement holds, every simple S is a quotient of R. Then one invokes the fact that if S is a quotient of M, then for any submodule N ⊆ M, S is a quotient of M or M/N. ]

Now the set of all rR for which rI = 0 and r(J/J’) = 0 is clearly $\left\{\begin{pmatrix} 0 & b \\ 0 & 0\end{pmatrix}\right\}.$ So this is J(R).

Exercise

Prove that the modules I and J/J’ are not isomorphic as R-modules. Thus R has exactly two simple modules up to isomorphism.

Finally, we prove the main

Theorem. If R is an artinian ring and J(R) = 0, then R is semisimple.

Proof

Let ∑ be the collection of maximal left ideals M ⊂ R. By definition of radical, $\cap_{M\in\Sigma} M = \text{rad}(R) = 0.$ We claim that there is a finite subset $\Phi \subseteq \Sigma$ such that $\cap_{M\in \Phi} M = 0.$

Indeed, for all finite $\Phi \subseteq \Sigma$, consider the left ideal $I_\Phi := \cap_{M\in \Phi} M.$

The collection of all such $I_\Phi$ has a minimal element, also written as $I_\Phi.$ If it is not zero, pick $x \in I_\Phi - \{0\}$ and since rad(R) = 0 there is an M∈∑ not containing x. But now $I_\Phi \cap M$ is properly contained in $I_\Phi,$ contradicting its minimality. This proves our claim.

Thus, the map $R\to \oplus_{M\in\Phi} R/M$ is injective. Since R/M is simple, the RHS is semisimple (as R-modules) and hence R is also semisimple. ♦

In conclusion:

Theorem. A ring R is semisimple if and only if it is artinian and J(R) = 0.

Exercise

Modify the above results to prove that the following are equivalent for an R-module M:

• M is semisimple and finitely generated.
• M is an artinian module and rad(M) = 0.
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