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 thelargest semisimple quotientof 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 thelargest semisimple quotientof 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

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 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 **R**^{3} 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 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*) = *J*^{2}*x* = *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

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.

**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 *x ^{n}* = 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. ♦