Jacabson Radical

Recall that the radical of the base ring R is called its Jacobson radical and denoted by J(R); this is a two-sided ideal of R. Earlier, we had proven that a ring R is semisimple if and only if it is artinian and J(R) = 0. Here we will demonstrate more properties of J(R).

For convenience, we will assume throughout this article that our base ring R is always artinian. Thus, we can take its quotient by the Jacobson radical to obtain S := R/J(R) in an attempt to make it semisimple by eliminating the radical.

Lemma. The collection of simple R-modules is precisely the collection of simple S-modules.

Proof

⇐ : in general, if I ⊂ R is any two-sided ideal, then a simple (R/I)-module is also a simple R-module.

⇒ : if M is a simple R-module, we saw earlier that J(R)M = 0, so M is in fact an R/J(R)-module. ♦

Proposition. S is semisimple, and is called the largest semisimple quotient of R because of the following universal property:

R → S is a surjective ring homomorphism to a semisimple ring;

if f : R → T is a surjective ring homomorphism to a semisimple ring T, then f factors through R → S (i.e. ker(f) ⊇ J(R)).

Proof

S is an artinian ring since any left S-module is also an R-module. Now it remains to show J(S) = 0. But J(S) is the intersection of all maximal left ideals of S, which corresponds to the intersection of all maximal left ideals of R containing J(R), which is precisely J(R). Thus J(S) = 0 and the first property holds.

For the second, T is semisimple as a T-module so it is a direct sum of simple left ideals of T. Since R→T is surjective, they are also simple left modules over R. On the other hand, every x in J(R) satisfies xM = 0 for any simple R-module M. Thus J(R)T = 0, and J(R) ⊆ ker(f). ♦

There is an analogous result for modules.

Proposition. Let M be an R-module; now N := M/(J(R)M) is the largest semisimple quotient of M due to the following universal property:

M → N is a module homomorphism to a semisimple module;

if f : M → P is a module homomorphism to a semisimple module, then f factors through N (i.e. ker(f) ⊇ J(R)M).

Proof

N is also a module over S := R/J(R), which we already know is a semisimple ring, so it is a sum of simple S-modules. Every simple S-module is a simple R-module, so N is a semisimple R-module. This proves the first statement.

For the second, since P is semisimple we have J(R)P = 0. So f(J(R)M) = J(R)f(M) ⊆ J(R)P = 0, and so J(R)M ⊆ ker f. ♦

But what about the radical of M? It turns out to be the same as J(R)M:

Lemma. If M is an R-module, J(R)M = rad(M).

Proof.

For any ring R, we always have J(R)M ⊆ rad(M). Indeed, the natural map

$M/\text{rad}(M) \longrightarrow \prod_{N\subset M \text{max}} (M/N)$

is injective and the image lies in a product of simple modules M/N. Since J(R)·(M/N) = 0 for each simple M/N, we also have J(R)M ⊆ rad(M) at the domain module.

Conversely, when R is artinian, the previous proposition says M/(J(R)M) is semisimple so it has zero radical, i.e. the intersection of all maximal submodules of M containing J(R)M is precisely J(R)M. So the intersection of all maximal submodules of M is contained in J(R)M and we get rad(M) ⊆ J(R)M. ♦

Before we proceed, let’s consider a concrete case.

Example

Let R be the set of 3 × 3 upper-triangular matrices with real entries $\begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \end{pmatrix}.$ We saw in the previous article that R has exactly 3 simple modules up to isomorphism. Explicitly:

$\begin{pmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{pmatrix} : \overbrace{\begin{pmatrix} u\\ 0 \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} au\\ 0 \\ 0 \end{pmatrix}}^A, \overbrace{\begin{pmatrix} 0 \\ v \\ 0 \end{pmatrix} \mapsto \begin{pmatrix} 0 \\ dv \\ 0 \end{pmatrix}}^B, \overbrace{\begin{pmatrix} 0 \\ 0 \\ w \end{pmatrix} \mapsto \begin{pmatrix} 0 \\ 0 \\ fw\end{pmatrix}}^C.$

We repeat that B and C are not submodules of the base ring. From the description of the simple modules, we get:

$J(R) = \left\{\begin{pmatrix} 0 & * & * \\ 0 & 0 & * \\ 0 & 0 & 0 \end{pmatrix}\right \}.$

So the largest semisimple quotient R/J(R) is just R × R × R. If M denotes the space R³ of column vectors, with R acting by left-multiplication, then rad(M) = J(R)M is just the space of (x, y, z) with z=0.

Further Properties

Finally, we wish to prove some rather important properties for artinian rings.

Theorem. J(R) is a nilpotent ideal of R.

Proof

Let I = J(R). Since R is artinian, the decreasing sequence $(I^n)_{n\ge 1}$ of two-sided ideals must terminate so eventually $I^{n+1} = I^n.$ Let J be this ideal; we get $J^2 = J.$ We need to show that J=0.

If not, let ∑ be the set of all left submodules K ⊆ J such that JK≠0. This is non-empty since it contains J, so pick a minimal element K. Choose x∈K such that Jx≠0, which gives J(Jx) = J²x = Jx ≠0. Thus Jx is in ∑ and since Jx ⊆ K, we have Jx = K by minimality of K. In particular, yx = x for some y in J; this gives (1-y)x = 0. Now apply the following lemma and we’re done. ♦

Lemma. For any y in J(R), 1+y is invertible.

Proof of Lemma.

Take the left ideal I := R(1+y) of R; we claim I = R. Indeed, if not, it must be contained in a maximal left ideal M. [ Proof: consider the collection of all left ideals containing I but not 1; Zorn’s lemma says there is a maximal left ideal here; since it doesn’t contain 1, it is proper and maximal. ] But this gives J(R) ⊆ M and since 1+y and y are both in M, we get 1∈M which is absurd.

Hence, R(1+y) = R so we have z(1+y) = 1 for some z in R. But now z = 1-zy is in 1+J(R) also, so there is a w such that wz = 1. This gives 1+y = wz(1+y) = w and so z is a two-sided inverse for 1+y. ♦

The next theorem is rather fascinating.

Hopkins-Levitzki Theorem. A left artinian ring is left noetherian.

Note

For commutative rings, we can say more: a ring is artinian ⇔ it is noetherian and all prime ideals are maximal.

Proof.

Let J = J(R) and consider the sequence of ideals

$R \supseteq J \supseteq J^2 \supseteq \ldots$

We just proved that J is nilpotent, so $J^n = 0$ for some n. Now each $J^{i-1}/J^i$ is a left-module annihilated by J so it is an (R/J)-module, i.e. $J^{i-1}/J^i$ is semisimple! Since R is artinian, so is each $J^{i-1}/J^i.$ A semisimple module is a direct sum of simple modules; if it is artinian, there are only finitely many terms in this direct sum, so it is noetherian too. Hence each $J^{i-1}/J^i$ is a noetherian R-module, and so is R. ♦

Next, we wish to strengthen the above theorem.

Theorem. The Jacobson radical J(R) is the unique maximal nilpotent ideal of R.

Proof

First we show that any nilpotent ideal I must be contained in J(R). Let x∈I. Since x is nilpotent, 1-x is a unit, for if xⁿ = 0 we can write:

$(1-x)(1+x+x^2 + \ldots+x^{n-1}) = \overbrace{1-x^n}^{=1} = (1+x+x^2+\ldots+x^{n-1})(1-x).$

If I is not contained in J(R), then it is not contained in some maximal left ideal M of R. This gives I + M = R, so we have 1 = x+y for some x∈I and y∈M. But we just saw that y=1-x is a unit, which contradicts the fact that M is proper!

Now we complete the proof.

Consider the collection ∑ of all nilpotent ideals of R; we know J(R)∈∑. Now if A and B are both in ∑, so is A+B. Indeed, AB lies in both A and B, so:

$A^m = B^n = 0 \implies (A+B)^{m+n} \subseteq A^m + B^n = 0$

since in each expansion of AABA…, either A occurs at least m times or B occurs at least n times. Since R is left artinian, it is also left noetherian, and ∑ has a maximal element. From the above, this maximal element must be J(R) and every nilpotent ideal is contained in it. ♦