6.1 The Jacobson Radical and Semisimple Artinian Rings

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In Theorem 5.9.5 of Section 5.9, we saw that rings all of whose left modules

were semisimple were essentially classified (as direct sums of matrix

rings). In the present section we shall define an ideal which serves as an

“obstruction” of the above condition.

Let R be a ring. Define the Jacobson radical of R by setting

J (R) = {r 2 R| rM = 0 for every irreducible R-module M}.

It is clear that J (R) is a left ideal of R. To see that it is also a right

ideal of R, let x 2 J (R) and let r 2 R. If M is an irreducible R-module,

then xrM _ xM = 0; since M was arbitrary, we conclude that xr 2 J (R).

Let is now denote by J 0(R) the set of all elements of R that kill every

irreducible right R-module. Thus J 0(R) is also a 2-sided ideal of R. We’ll

see momentarily that J 0(R) = J (R).

Here’s our main characterization of J (R).

Theorem 6.1.1 The following ideals in R are identical.

(1) J (R).

(2) \MM, where M ranges over all maximal left ideals of R.

(3) [II, where I ranges over all left ideals of R such that 1 + I consists

entirely of units.

149

150 CHAPTER 6. RING STRUCTURE THEORY

(4) {r 2 R|1 + arb is a unit in R for all a, b 2 R}.

(5) J 0(R).

(6) \MM, where M ranges over all maximal right ideals of R.

(7) [II, where I ranges over all right ideals of R such that 1 + I consists

entirely of units.

An element r 2 R is said to be nilpotent if rn = 0 for some positive

integer n. An ideal I _ R is called nil if every element of I is nilpotent.

Finally, an ideal I _ R is nilpotent if In = 0 for some positive integer n.

Note that every nilpotent ideal is nil.

Example 1. Let F be a field and let

R =

__

a b

0 c

_

| a, b, c 2 F

_

.

Now set

I =

__

0 b

0 0

_

| b 2 F

_

.

Note that I is nilpotent.

Example 2. Let p be a prime, let n be a positive integer, and let R =

Z/(pn). For any positive integer m, the ideal pmR is nilpotent, hence

nil. (See Exercise 8, below.)

Example 3. Here is an example of an ideal I in a ring R such that I is nil

but not nilpotent. Let F be a field, and set R = F[x1, x2, x3, . . .], a

polynomial ring in an infinite number of indeterminates. Let A _ R

be the ideal generated by {x21

, x32, x43

, . . .}, and set ¯R = R/A. If r 2 R

let ¯r 2 ¯R denote the image of r in ¯R under the canonical map R ! ¯R.

If ¯I _ ¯R is the ideal (¯x1, ¯x2, . . .), then one easily checks that ¯I is nil.

On the other hand, if n is a positive integer, note that 0 6= ¯xnn

2 ¯I,

and so ¯I is not nilpotent.

Proposition 6.1.2 If I _ R is a nil left ideal, then I _ J (R).

Corollary 6.1.2.1 If I _ R is a nilpotent left ideal, then I _ J (R).

Lemma 6.1.3 Let R be a ring and let I be a non-nilpotent minimal left

ideal of R. Then I contains a non-zero idempotent.

6.1. THE JACOBSON RADICAL 151

Corollary 6.1.3.1 Let I _ R be as above. Then R = I _ I0, for some

left ideal I0 _ R. More generally, if J is a left ideal of R, and if I _ J is a

non-nilpotent minimal left ideal of R, then J = I _ J0, for some left ideal

J0 _ J of R.

Proposition 6.1.4 (Nakayama’s Lemma) Let M be a finitely generated

Rmodule.

Then J (R)M = M if and only if M = 0.

The ring R is called left Artinian if the left R-module R is an Artinian

module. Similarly we can define what it means for R to be right Artinian,

left Noetherian and right Noetherian.

We now have the following.

Theorem 6.1.5 Let R be a left Artinian ring. Then the Jacobson radical

J (R) is a nilpotent ideal.

A ring R is called semisimple if J (R) = 0. Note that this is different

from saying that the left R-module R is semisimple. For example the reader

can easily check that Z is a semisimple ring, but is certainly not a semisimple

module. Here’s the relationship between the two concepts of semisimplicity:

Theorem 6.1.6 Let R be a left Artinian ring. Then the following are

equivalent.

(i) R is a semisimple ring.

(ii) R is a semisimple left R-module.

Corollary 6.1.6.1 (Wedderburn’s Theorem) A semisimple left Artinian

ring is a direct sum of matrix rings over division rings.

Corollary 6.1.6.2 A semisimple left Artinian ring is also right Artinian.

Finally, we have the following mildly surprising result.

Theorem 6.1.7 (Hopkin’s Theorem) A left Artinian ring is left Noetherian.

Exercises 6.1

152 CHAPTER 6. RING STRUCTURE THEORY

1. Consider the infinite matrix ring R = M1(F) over the field F, which

consists of matrices with countably many rows and columns, but such

that each matrix has only finitely many non-zero elements in any given

row or column. Show that in R, there are elements that are left (right)

invertible, but not right (left) invertible. (Hint: Let A be the matrix

having 1’s on the super-diagonal, and 0’s elsewhere. Let B be the

matrix having 1’s on the sub-diagonal and 0’s elsewhere. Note that

AB = I.)

2. Let R be a ring and assume that the element a 2 R has a unique left

inverse. Prove that a is invertible, i.e., the left inverse of a is also the

right inverse of a.

3. Let a 2 R and assume that a has more than one left inverse. Prove that

in fact a has infinitely many left inverses (thus R is infinite). (Hint:

If a has exactly n left inverses b1, b2, . . . , bn, set di = b1 + 1 − abi, i =

1, 2, . . . , n. Note that the elements di are pairwise distinct and are also

left inverses for a. If di = b1 for some i, obtain a contradiction.)

4. Let R be a ring such that for all 0 6= a 2 R, Ra = R. Prove that R is

a division ring.

5. Let R be a ring without zero divisors such that R has only finitely

many left ideals. Prove that R is a division ring. (Hint: Assume that

0 6= a 2 R and Ra 6= R. Look at the sequence Ra _ Ra2 _ · · ·.)

6. Let R be a ring and let L be the intersection of all non-zero left ideals

in R. If L2 6= 0, then R is a division ring. (Hint: By Lemma 6.1.3, we

have L = Re, where e is a non-zero idempotent of L. Next, if xe 6= x

for some x 2 R, then xe − x is in the left annihilator AnnR(e) =

{r 2 R | re = 0} of e. Since AnnR(e) is also a left ideal of R, we

get L _ AnnR(e), which is a contradiction. Therefore xe = x for all

x 2 R, so L = R. This implies that Ra = R for all 0 6= a 2 R; apply

Exercise 4.)

7. Assume that the ring R has no non-zero nilpotent elements. Prove that

every idempotent of R is contained in the center of R (i.e., commutes

with every element of R).

8. Let n be a positive integer, and let R = Z/(n). Describe the nilpotent

ideals in R.

6.1. THE JACOBSON RADICAL 153

9. If R is a ring, prove that J (R) contains no non-zero idempotents.

10. Let R be the ring of all continuous real-valued functions on the interval

[0, 1]. Prove that J (R) = 0.

11. Let R be a ring. Prove that J (R/J (R)) = 0.

12. Let R be a left Artinian ring, and let I _ R be a nil ideal. Prove that

I is actually nilpotent.

13. Let F be a field, and let R be the ring

R = =

8<

:

2

4

a1 a2 a3

a4 a5 a6

0 0 a7

3

5| ai 2 F

9=

;.

Compute J (R).

14. Let R be a left Artinian ring, and let I _ R be a non-nilpotent left

ideal. Prove that I contains a non-zero idempotent.

15. Let R be an Artinian ring. Prove that the following conditions are

equivalent.

(a) R is local, i.e., it has a unique maximal ideal.

(b) R contains no non-trivial (i.e. 6= 1) idempotents.

(c) If N is the radical of R, then R/N is a division ring.

Chapter 7 Tensor Products

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