Radical ideals and nil radicals can be defined for commutative or noncommutative rings. The definitions and theorems are more complicated in the noncommutative world, so you may want to look at radicals in commutative rings first, then return to this chapter.

You've probably seen these definitions before, but let's run through them again.

An element x in a ring R is nilpotent if some power of x = 0.

A left ideal H is nil if every x in H is nilpotent. The entire left ideal H is nilpotent if Hn = 0. Here Hn is shorthand for H*H*H… n times. Remember that the product of two left ideals is another left ideal. After finitely many multiplications of H, the result is 0.

If Hn = 0 then xn = 0 for every x in H. Thus a nilpotent ideal is a nil ideal. The converse does not hold, even for commutative rings, and even if the exponent is bounded. Start with the integers mod p and adjoin infinitely many indeterminants x1 x2 x3 … Set xip = 0 for each xi. A product of these indeterminants, with any coefficient from 1 to p-1, when raised to the p, is 0. By the frobenius homomorphism, every polynomial (with no constant term) raised to the p is 0. The ideal H generated by the indeterminants is nil, with a uniform exponent of p, yet Hn includes x1x2x3…xn, which is nonzero. Saying H is nilpotent is significantly stronger than saying H is nil.

Assume G and H are nilpotent left ideals, with exponents k and l, and consider G+H. Raise a+b in G+H to the k+l-1. Remember that a and b may not commute, so the binomial theorem does not apply. Expand the product, and each string contains at least k instance of a, or l instance of b. This is k elements from G multiplied together, or l elements from H multiplied together. Each string drops to 0. Therefore G+H is nilpotent.

If G is nilpotent and H is nil, and bl = 0, then raise a+b to the kl. A string might begin with b, but by position l we have to see a, else the string is 0. By position 2l there is another a, and so on up to kl, but this is k elements of G multiplied together. The sum of a nilpotent left ideal and a nil left ideal is nil.

Kothe's conjecture says the sum of two left nil ideals is a left nil ideal. This remains an open question. More on this below.

By induction, the finite sum of nilpotent left ideals is another nilpotent left ideal. However, H could be generated by infinitely many nilpotent ideals, without being nilpotent. Let the ith ideal be generated by the ith indeterminant xi in the earlier example. H will be a nil ideal however, since everything in H lives inside a finite sum of nilpotent ideals.

Move on to two sided ideals. If an ideal is generated by nil ideals, it is nil. Start with a nil ideal H and a nil left ideal J. Take a from H and b from J, and look at a+b mod H. This is the same as b mod H, hence (a+b)n is 0 in R/H. Pulling back, (a+b)n lies in H and is nilpotent, hence a+b is nilpotent.

Induction makes a finite sum of nil ideals nil, and any x in an infinite sum of nil ideals is a finite sum, contained in the sum of finitely many nil ideals, and is nilpotent. Thus the sum of nil ideals is nil.

The sum of all the nil ideals in R is itself a nil ideal - the largest nil ideal of R. This is called the upper nil radical of R, written upnil(R).

Note that x is in upnil(R) iff the principal ideal generated by x is nil.

Let H be a nil ideal that is finitely generated as a left ideal, and assume the generators lie in the center of R. Every x in H is a linear combination of these generators, with coefficients on the left. Multiply n such elements together, and expand the product of sums into a sum of products. In each term, move the generators to the right. If n is sufficiently large, at least one of the generators will become 0, since all the generators are nilpotent in H, and commute with each other. The terms drop to 0, the product is 0, Hn = 0, and H is nilpotent. The order of H is determined by the orders of the generators.

A nilpotent ideal could be infinitely generated. Adjoin infinitely many indeterminants to Z, and set all products of indeterminants to 0. Let the indeterminants generate H, wherein H2 = 0. An infinite ascending chain of ideals shows H is not finitely generated.

Recall that a ring is prime if 0 is a prime ideal, and semiprime if 0 is a semiprime ideal.

If H is an ideal in a semiprime ring then its left and right annihilators are the same. Let c*H = 0, whence (Hc)2 = 0. Since 0 is a semiprime ideal, Hc = 0.

Assume R is reduced, i.e. R has no nilpotent elements, and convert the above proof to elements. Let c annihilate x from the left, whence cx = 0. Thus (xc)2 = 0, and xc = 0.

Let R be a prime reduced ring, and let xy = 0. Since x kills y from the left it kills y from the right, hence yx = 0. The same holds for any left multiple of x. The left ideal generated by x kills y from the left, and the right. The right ideal generated by x kills y from the right, and the left. Let x generate the ideal G and let y generate H. G kills y from the left or the right. By associativity, G kills H from the left or the right. Thus gh = 0, and since 0 is prime, either G or H is 0. Thus x or y is 0, and R is a domain, free of zero divisors.

Let S be a nonempty subset of a ring R. If, for every x and y in S, there is w in R, such that xwy is in S, then S is an m-system. The "m" is suppose to remind you of multiplicatively closed sets. In fact a multiplicatively closed set is an m-system - set w = 1.

Let P be an ideal in R. Recall that P is prime iff, for every x and y, xwy in P for all w in R implies either x or y lies in P. This is the xRy test. Therefore P is prime iff the complement S = R-P is an m-system. This is analogous to the commutative case: P is prime iff S is multiplicatively closed.

If an m-system S misses an ideal H, use zorn's lemma to find a maximal ideal P missing S, and containing H. Suppose ideals A and B do not lie in P, but AB does. Thus P+A intersects S in some element p1+x. Similarly P+B meets S in p2+y. For some w, (p1+x) * w * (p2+y) lies in S. Expand this product, p1wp2 + p1wy + xwp2 + xwy, and each term lies in P, hence the product lies in P, which is a contradiction. Therefore P is prime. This is analogous to the commutative case: a maximal ideal missing a multiplicatively closed set is prime.

 P m-sys H
There is an n-system, similar to an m-system. The "n" in n-system stands for nilpotent. For any x in an n-system, there is w such that xwx lies in S. The complement of an ideal C is an n-system iff C is semiprime. Use the xRx test for primality.

If an ideal H misses an n-system S, drive H up to a maximal ideal missing S, giving a semiprime ideal. The proof is as above: (p1+x) * w * (p1+x) is in both P and S.

 S n-sys H

Let 1 define an m-system, and every maximal ideal misses this m-system. Thus every maximal ideal is prime (and semiprime), just as it was in the commutative world. Yes, we proved this earlier , using the fact that S (equal to 1) was multiplicatively closed.

There is an m-system inside every n-system. Let x1 be an element in a nonempty n-system. There exists a w1 with x1w1x1 in S. Let x2 = x1w1x1. Let x3 = x2w2x2, and so on, building an infinite sequence. Verify x1 x2 x3 etc is an m-system. For i ≤ j, xj is in the left ideal generated by xi, and in the right ideal generated by xi. If xj is on the right, write xiy = xj. Set w = ywj, and xiwxj = xj+1. If xj is on the left, find y such that xj = yxi, and set w = wjy. Once again, xjwxi = xj+1. That establishes the m-system.

Let H be an ideal of R. As a set, let rad(H) be the elements x, such that every m-system containing x intersects H.

If a prime ideal P contains H and misses x, x is in the m-system R-P, and is not in rad(H), hence rad(H) lies in every P containing H. Conversely, if x is in every prime containing H, i.e. the intersection of the primes containing H, try to embed x in an m-system S missing H. Drive H up to a maximal ideal missing S, which is prime, and x is not in every prime containing H. Therefore rad(H) is the intersection of the prime ideals containing H, and rad(H) is an ideal. This definition agrees with the definition of rad(H) in a commutative ring.

If H is proper, it misses the m-system 1, and can be raised to a maximal, prime ideal in R containing H. Thus rad(H) is always a proper ideal.

A minimal ideal need not be prime. Let P2 generate H inside the integers mod p3.

H is a radical ideal iff it is semiprime. Let H be semiprime. We know H lies in rad(H), so suppose x is in rad(H), but not H. Now x is in an n-system missing H, namely the complement of H, and x seeds an m-system lying entirely within the n-system. Thus x lies outside a prime containing H, and x does not lie in rad(H) after all. We have H = rad(H), and a semiprime ideal is a radical ideal.

Conversely, let H = rad(H), the intersection of all prime ideals containing H. Since a prime ideal is semiprime, H is the intersection of semiprime ideals. The intersection of semiprime ideals is semiprime; A2 in all of them implies A is in all of them. Therefore H is semiprime.

This is compatible with commutative rings. H is semiprime iff xn in H implies x is in H, iff H is a radical ideal.

Let H be any proper ideal of R, and let J be the intersection of semiprime ideals containing H, hence J is itself semiprime. Since prime ideals are semiprime, rad(H) contains J contains H. As shown above, J = rad(J), so rad(H) cannot be larger than J. Therefore rad(H) = J.

In summary, rad(H) is the least semiprime ideal containing H, or the intersection of semiprime ideals containing H, or the intersection of prime ideals containing H. This is valid for commutative or noncommutative rings.

Recall that the upper nil radical of a ring R, written upnil(R), is the largest nil ideal of R. The lower nil radical is the intersection of all prime ideals, written lownil(R). This is a radical ideal, specifically, rad(0). It is also a semiprime ideal, the smallest semiprime ideal containing 0, the smallest semiprime ideal of R, and the intersection of all the semiprime ideals of R. Note that R/lownil(R) is a semiprime ring, since 0 in the quotient is a semiprime ideal by correspondence.

If x lies in lownil(R) then the powers of x form an n-system that intersects 0. (Otherwise 0 could rise to a semiprime ideal missing x.) Thus x is nilpotent, and lownil(R) is a nil ideal. This is contained in upnil(R), the largest nil ideal, which is contained in jac(R) (holding all the nil ideals). If R has no nil ideals, as when R is reduced, or jacobson semisimple, upnil(R) = lownil(R) = 0.
 R jac(R) upnil(R) lownil(R) 0

If R is commutative, all the nilpotent elements form a nil ideal, which becomes the largest nil ideal of R. This in turn is rad(0). Thus lownil(R) = upnil(R), and we simply write nil(R). Such a ring is reduced iff nil(R) = 0, iff R is a semiprime ring. A noncommutative ring however could be semiprime without being reduced. Adjoin x and y to the field of order 2, where x and y do not commute, and set x2 = 0. x is the only nilpotent element, and it does not participate in a nil ideal. The upper and lower nil radicals are both 0, yet x floats above.

Let R be left artinian, whence jac(R) is nilpotent. Since Jn lies in lownil(R), and lownil(R) is semiprime, J lies in lownil(R). Thus jac(R) = upnil(R) = lownil(R).

You have to look around for a while to find a ring where upnil(R) properly contains lownil(R). Add 1 to this ring, and jac(R) = upnil(R), yet 0 is a prime ideal, so lownil(R) = 0.

Let S be the n×n matrix ring over R.

Recall that the ideals of S correspond 1 for 1 with the ideals of R. In fact, the matrices that are 0, except for the upper left entry, implement the correspondence. The map from the ideals of R onto the ideals of S respects containment, intersection, sum, and product. Therefore prime and semiprime ideals correspond.

Since the lower nil radical is the least semiprime ideal, lownil(S) is the matrix ring over lownil(R).

More generally, the least semiprime ideal in R containing H extends to the least semiprime ideal in S containing the matrices of H. In other words, the extension of the radical equals the radical of the extension.

If H is a nilpotent ideal of R, with Hn = 0, then expand the product of n matrices over H. Each entry is a sum of products from Hn; each entry is 0, and the matrix is 0. Conversely, if the product of any n matrices over H is 0, then look at the matrices that are 0 except for the upper left, and Hn is 0 in R. Nilpotent ideals correspond, having the same exponent.

A nil ideal in S pulls back to a nil ideal in R, by restricting to the upper left entry, but the converse is not clear.

See jacobson radicals for a proof that jac(R) and jac(S) correspond.

If R is commutative, and P is a prime ideal in R, then the polynomials with coefficients in P form a prime ideal in the ring of polynomials R[x]. Similarly, semiprime in R extends to semiprime in R[x], and extention and radical commute. If H is any proper ideal of R, the polynomials with coefficients in rad(H) form the radical ideal of the polynomials with coefficients in H. You can review the proof if you like; what follows is largely a recapitulation of that logic, using the more general definition of a prime ideal in a noncommutative ring. It isn't particularly harder, I just talk about xRy in P instead of xy in P.

Let S = R adjoin arbitrarily many indeterminants t1 t2 t3 etc, and assume S is a prime ring. As usual, the indeterminants commute with R and with each other. Select x and y in R, and assume xwy = 0 for every w in R. Now xey = 0 for any polynomial e in S. Thus x or y is 0, and R is a prime ring. (Indeterminants need not commute with each other for this part.)

Conversely, let R be prime, and assume fwg = 0 for all w in S. Restrict to R adjoin finitely many indeterminants, specifically those exhibited by f and g. The product fwg remains 0, and if the ring is prime at this finite level, either f or g is 0, and S is a prime ring. It is enough to prove the theorem for finitely many indeterminants, and by induction, it is enough to adjoin one indeterminant t.

Select w from R, and fwg is still 0. The lead coefficient of fwg is 0 for every w in R, hence one of the two lead coefficients is 0. Either f or g is 0, and S is a prime ring.

An analogous proof shows R is semiprime iff S is semiprime.

Let H be an ideal in R, and extend it into S, i.e. the polynomials with coefficients in H. Now R/H is prime iff S/H is prime. As ideals, H is prime iff its extension into S is prime. An analogous proof shows H is semiprime iff its extension is semiprime.

Let H be a prime ideal in R that extends to a prime ideal C in S. Let D be a prime ideal properly contained in C. Thus D does not contain all of H, as that would generate all of C. In fact D intersects R in an ideal J, properly contained in H. Is J a prime ideal in R? Assume there are elements a and b in R, where each awb lies in J. Extend w to be any polynomial in S. Now every coefficient of awb lies in J, hence awb is spanned by J, and lives in D. Therefore a or b lies in D, and in J, whence J is prime. Ratchet C down to a smaller prime ideal D, and D intersects R in a prime ideal below H. The extension of a prime ideal H in R gives the least prime in S containing H. A similar proof shows the extension of a semiprime ideal in R gives the least semiprime ideal in S that contains the original ideal in R.

Let C = lownil(S), and let H = lownil(R). Remember that H is the least semiprime ideal in R, and its extension into S is semiprime, and contains C. If C is a smaller semiprime ideal then C intersects R in a semiprime ideal below H. This is a contradiction, hence C is the extension of H. In other words, the extension of the lower nil radical equals the lower nil radical of the polynomial ring.

Generalize this to show radical and extension commute. Let H = rad(J), the smallest semiprime ideal containing J. Extend J into S and let C be its radical, the least semiprime ideal containing J. The extension of H is semiprime, and contains J, and C. If C is a smaller semiprime ideal, still containing J, then C intersects R in a semiprime ideal below H, and still containing J. This is a contradiction, hence C is the extension of H. In other words, the extension of the radical equals the radical of the extension.

Let H be a minimal left ideal. Either H is a summand R*e for some idempotent e, or H2 = 0.

Suppose the latter is not the case, hence there is some x in H with Hx nonzero. Hx is a nonzero left ideal. By minimality, Hx = H. Find e in H with e*x = x.

Let B be the left annihilator of x contained in H. B is a left ideal that misses e, hence B = 0. Now e2-e kills x and lies in H, hence it lies in B, and e2-e = 0. In other words, e2 = e. This makes e an idempotent.

Since R*e is a nonzero submodule of H, it equals H, and H is a principal left ideal generated by the idempotent e.

Set f = 1-e, and R*e and R*f span R. We've seen this before, but let's run through it again. e2 = e, f2 = f, ef = fe = 0. Suppose y*f = x*e. Multiply by f on the right and yf = 0; multiply by e on the right and xe = 0. The modules are disjoint, and since multiplication by 1 (e+f) covers R, R is the direct product of these two modules. H is a summand of R.

As you recall, semisimple is equivalent to jacobson semisimple and dcc. (You only need dcc on left principal ideals to run the reverse direction.) If jac(R) is 0 then lownil(R) = 0, and R is semiprime. This leads to the stronger statement: semisimple iff semiprime and dcc.

Let R be semiprime, with dcc on its principal left ideals. Given a left ideal H, let x generate a principal left ideal inside H. Keep extracting smaller principal left ideals; but this can't go on forever. It stops when e generates H, where H is a minimal left ideal.

If H2 = 0 then H = 0, since R is semiprime. By Brauer's lemma, we can assume e is idempotent, and H becomes a summand of R.

Since H is minimal it is a simple left R module. We have pulled a simple module off of R. Call this U1, having summand M1. Within M1 is another simple module U2, having summand M2. Let V2 = M1 intersect M2. As a left R module, R is the direct sum of U1, U2, and V2. This is the same construction you saw before. V2 contains U3, spinning off V3, and so on, but by dcc this cannot continue forever, so R is the finite direct product of simple modules, and R is semisimple.

The element x generates a left nil ideal in R iff it generates a right nil ideal. Assume the former, and write (yx)n = 0. Thus (xy)n+1 = 0.

If every left multiple of x is nilpotent then the same is true for the left multiples of yx. In other words, the elements that generate nil left ideals can be scaled.

The sum of all left nil ideals is the same as the sum of all right nil ideals, i.e. the sum over all x that generate one sided nil ideals. This set is an ideal, denoted onenil(R). Since onenil(R) contains every nil ideal, it contains upnil(R). The only catch is, nobody knows if it is nil. If it is, onenil(R) = upnil(R). This is equivalent to asserting every one sided left nil ideal lies in upnil(R).

Since every left nil ideal lies in jac(R), jac(R) contains onenil(R).

 R jac(R) onenil(R) upnil(R) lownil(R) 0

Kothe's conjecture states that the sum of two left nil ideals is another left nil ideal. Let's relate this to onenil(R).

If all left nil ideals lie in upnil(R) then the sum of any set of left nil ideals lies in upnil(R) and is nil.

Conversely, if the sum of finitely many left nil ideals is nil then the same holds for infinitely many left nil ideals, for each element lives in the sum of finitely many left nil ideals. The sum of all left nil ideals is a nil ideal, namely onenil(R), which lives in, and contains, upnil(R), hence they are equal.

Another equivalent condition states every left nil ideal generates a nil ideal. If true, the sum of two left nil ideals is contained in the sum of the two generated nil ideals, which is nil, hence kothe's conjecture holds.

Conversely, if every left nil ideal is in upnil(R) then the two sided ideal generated is in upnil(R), and is nil.

A ring that satisfies these conditions is called kotherian. If Emmy Noether can lend her name to an adjective, noetherian, why not Kothe? Note that a commutative ring is kotherian, with lownil(R) = upnil(R) = onenil(R).

If H is a left nilpotent ideal, the ideal generated by H is nilpotent with the same exponent. An element e in the ideal of H can be represented as a linear combination of elements of H with coefficients from R on the right. Raise e to the nth power, and expand it into a sum of products. Each string is the product of n elements from H, times a coefficient on the right. This is 0, hence en = 0, and the ideal has order n. The one sided nilpotent ideals all live in upnil(R).

If H is a left nil ideal and every finitely generated left ideal inside H is nilpotent then H generates a nil ideal. Represent e as above, and note that e is in the span of a finitely generated nilpotent left ideal. Thus e lives in a nilpotent ideal, en = 0, and the ideal generated by H is nil.

Let H be a right nil ideal in a ring R. Let V be the set of right ideals that annihilate various elements of R from the right. Thus V contains R, since 0*R = 0, and V contains 0, since 1*0 = 0.

If V satisfies acc, then H lives in lownil(R). Furthermore, H contains a nilpotent right ideal. This theorem courtesy of Levytsky.

Suppose H is a nonzero right nil ideal, not contained in lownil(R). Restrict V to right ideals that kill elements of H that are not in lownil(R). V remains acc, free of infinite ascending chains. Choose x so that its right annihilator J is maximal in V. Since x is nilpotent, j contains some power of x, and is nonzero.

For any w in R, let y = xwx, and note that the right annihilator of x lies in the right annihilator of y. Also, xw lies in H, and is nilpotent.

If xw = 0 then y = 0, which is in lownil(R).

If xw is nonzero, let (xw)n = 0. Thus w*(xw)n-2 kills y, but not x. This creates a right annihilator larger than J. By maximality, y lies in lownil(R).

For every w, xwx lies in lownil(R), and since lownil(R) is a semiprime ideal, x lies in lownil(R) after all. Therefore H lies in lownil(R).

All the right nil ideals combine to make onenil(R), which lives in lownil(R), which lives in upnil(R), which lives in onenil(R). All three ideals are equal. As a corollary, R is kotherian.

By symmetry, we can make the same claim if the left annihilators satisfy acc. If R is left or right noetherian, onenil(R) = upnil(R) = lownil(R).

Return to H, our right nil ideal, which now lives in lownil(R). Let x lie in H, having a maximal right annihilator. If xwx is nonzero then y = xwx implies a larger right annihilator. Therefore xwx = 0 for all w in R. The square of the right ideal generated by x is 0. This is a principal right nilpotent ideal inside H.

Let R be noetherian with respect to its two sided ideals, and let M be a maximal nilpotent ideal. If H is another nilpotent ideal then H+M is nilpotent, and is contained in M, hence H is contained in M. In other words, M is the largest nilpotent ideal in R.

If H2 lies in M then H is nilpotent, and M already contains H. Therefore M is a semiprime ideal, containing lownil(R), which is the smallest semiprime ideal.

With Mn = 0, Mn lies in lownil(R), and with lownil(R) semiprime, M lies in lownil(R). Therefore M = lownil(R). If R is noetherian on its ideals, or left noetherian, or right noetherian, lownil(R) is the largest nilpotent ideal.

Finally, assume R is noetherian on its ideals, and noetherian on its left or right annihilators. (This is implied if R is left or right noetherian.) Combine the earlier results to assert onenil(R) = upnil(R) = lownil(R) = the largest nilpotent ideal. Every nil ideal, or nil left/right ideal, is nilpotent, with a maximum exponent that applies across the board.

The brown macoy radical, denoted brown(R), is the intersection of the maximal two sided ideals of R. Clearly brown(R) = jac(R) in a commutative ring.

Suppose y is in jac(R), but not in brown(R). Thus y is not in one of the maximal ideals M of R. M and jac(R) span 1. Write x ∈ M + z ∈ jac(R) = 1. Thus x = 1-z, yet 1-z is a unit, and x cannot live in M. This is a contradiction, therefore brown(R) contains jac(R).

 R brown(R) jac(R) onenil(R) upnil(R) lownil(R) 0

At this point you're probably looking for a ring where the brown radical properly contains the jacobson radical.

Let T be the ring of endomorphisms on a countably infinite dimensional K vector space. These are the infinite matrices that we saw earlier. Excise each column in turn to find a countable set of left maximal ideals, whose intersection is 0. Thus jac(T) = 0.

In that section I defined a subring S: the endomorphisms whose images are finite dimensional. These are the matrices that are 0 beyond some column. Verify that S is a two sided ideal inside T.

In that section I showed that every matrix brings in all of S. Thus S is part of every nonzero ideal, and part of every maximal ideal. Furthermore, T/S is a simple ring, hence brown(T) = S.