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.
⇐ : 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)).
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).
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).
For any ring R, we always have J(R)M ⊆ rad(M). Indeed, the natural map
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.
Let R be the set of 3 × 3 upper-triangular matrices with real entries We saw in the previous article that R has exactly 3 simple modules up to isomorphism. Explicitly:
We repeat that B and C are not submodules of the base ring. From the description of the simple modules, we get:
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.
Finally, we wish to prove some rather important properties for artinian rings.
Theorem. J(R) is a nilpotent ideal of R.
Let I = J(R). Since R is artinian, the decreasing sequence of two-sided ideals must terminate so eventually Let J be this ideal; we get 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.
For commutative rings, we can say more: a ring is artinian ⇔ it is noetherian and all prime ideals are maximal.
Let J = J(R) and consider the sequence of ideals
We just proved that J is nilpotent, so for some n. Now each is a left-module annihilated by J so it is an (R/J)-module, i.e. is semisimple! Since R is artinian, so is each 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 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.
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:
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:
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. ♦