THE KRULL RADICAL , k-PRIMITIVE RINGS , AND CRITICAL RINGS

We generalize results on the Krull radical, k-primitive rings, and critical rings from rings with identity to rings which do not necessarily contain identity..


R. TUCCI
primitive rings.We list the main properties of these rings and generalize slightly a theorem on these rings (Prop. 3.4).We finally turn our attention to crit- ical rings, which are closely related to k-primitive rings.Necessary and sufficient conditions are given for a critical ring to be a domain, and those critical rings which are not domains are completely characterized.In what follows the letter R denotes an associative ring which does not necessar- ily contain identity.An R-module is a right R-module; usually we will simple call this module M.
Let Z denote the integers.We define R I (r, n) Ir R, n Z}, where addition is componentwlse and multiplication is given by (r, n) (r', n') (rr' + nr' + n'r, nn').This is just the usual ring with identity in which R +/-s embedded.For nota- tional simplicity, we identify R with the subring (R, 0) of R I to which R is iso- morphic.All modules over R I are unital, so that every R-module M can be considered an Rl-mOdule if we define m(r, z) mr + mz for all m M, (r, z) R I. Conversely, any Rl-modle M can be considered an R module with scalar multiplication defined by mr m(r, 0) for all m M, r R. Krull dimension for an R-module M is defined as in [2] and is denoted K dim M, or sometimes K dim M R A familiarity with the results and M of this paper is assumed.Note that for any R-module M, K dim K dim i is a k-critical R-module if and only if M is a k-critical Rl-mOdule.Finally, E (M) denotes the injective hull of this module M. 2.
THE KRULL RADICAL.R As in [i] we say that a right ideal H of a ring R is k-co-crltical if is a k-critical R-module.A right ideal H of R is n_-modular if there exist e R, 0 n Z, such that er nr H for all r R. If n I, then we call H modular in accordance with the usual terminology.A right ideal which is either maximal modular, 1-co-critical and n-modular, or k-co-critical, k > 2, is called a special co-crltical r_ht ideal of R. The Krull radical of R, denoted K(R), is defined to be the inter- section of all the speical co-critical right ideal of R, if any exist; if there are results of [l], we first prove that K(R) K(RI).
LEMMA 2.1 Let H be a right ideal of R, H R, and let H I {(e, -n) E R I er-nr E H for all r E}.

Then
(1) H I is the unique right ideal of R l which is maximal with respect to the property that H I n R H; (2) H is the n-modular if and only if H I R.
PROOF (i) It is routine to verify that H I is a right ideal of R 1.
The uniqueness of H I follows from the observation that H I {x R I x R __c H}.
(2) LEMMA 2.2 Let M be a trivial R-module; i.e., mr 0 for every m M, r E R.
If K dim M k exists, then k < I.
PROOF Since M is a trivial R-module, its R-module struc{ure is the same as its structure as an abelian group i.e., as a Z-module.By [2, Cor. 4 PROOF By [3, p. ii, Thm. 2]and by definition of the Krull radical, K(R I) J(R I) J(R) R. Thus, K(R I) n (H 1 n R), where C is the set of co- HIC critical right ideals of R I. We will show that the set of special co-crltical right ideals of R coincides with the set of right ideals of the form H I n R, where H I C.
Since J(R) J(RI) we do not need to consider the case where H I is a maximal right ideal of R I.
Suppose that R has some special co-critical right ideals.Let H be a special k- co-critical right ideal of R, k > 0, and let H 1 be as in Lemma 2.1.We first deter- R I mine K dim i" By [2, Lemma i To show.H 1 is co-critical, let K 1 be any right ideal of R 1 which properly con- tains H I. Then R o K I properly contains H because of the way H 1 is defined.Repeat- ing the argument we used to find K diml gives us that Conversely, suppose that H I is a k-co-critical right ideal of R I, k > 0, and assume H I R (if there is no such H I then R has no special co-critical right ideals 2 then H is a special co-critical right ideal of R. Suppose k i.Then H I R; for if H I R, then there is an onto map from R I R 1 R I H to -.Since both modules have Krull dimension i, andI is critical, we must have R 1 R 1 R I i =-.But then1 R 0, which implies R H 1 and this in turn implies that R R H I 0 R H, contradicting the fact that K dim i.Thus, H I R. We must have, then, that H 1 contains some (e, -n) R with n + 0, so for every r R, (e, -n)r er nr H 1 0 R H. Hence H I is n-modular, and therefore special.
Suppose now that R has no special co-critical right ideals.Then K(R) R by R 1 definition.Since-is 1-critical, R is a co-critical right ideal of R I. Every other co-critical right ideal H I of R I contains R; for, if R H I then H I R is a special co-critical right ideal of R, contradiction.Therefore, K(R I) R K(R).
This completes the proof.COROLLARY 2.4 (i) K(R) is the set of elements of R which annihilate every critical right R-module.
The next result shows that K(Rn) (K(R)) n, where Rn is the ring of n x n matrices over R. If R has identity, then Eli denotes the matrix wlth i in the (i, j) position and zeroes elsewhere.
LEMMA 2.5.Let R be a ring with identity, and let H be a right ideal of R.
Take H (1) to be the set of all matrices in R whose i th row has entries from H and R whose other entries are arbitrary.Then (i) is a critical -module if and only if For simplicity, asse that i i.Note that Htj conslstsR of matrices n whose only non-zero row is the first.Thus, any smodule S of (i) can be itten H(1) Ejj H-for any I j n, and Ejj consists of matrices with nonzero entries in the (l, j) position i.e., from N and zeroes elsewhere.But then for R R This implies that Nj H(1) N k.Since j and k are arbitrary, we have that N 1 Nn.Call this set N. It is routine to check that R N is an R-submodule of .Thus, there is a i-i onto order preseing map f from the R n R
LEMMA 2.6.Let R be a ring with identity.If M is a cyclic critical R-module, (+/-) then there is a co-critical right ideal H = R as in Lemma 2.5 such that M is n isomorphic to R H(i)

PROOF
Since M is cyclic, we can find a matrix A M such that M AR We show n first that there is an integer j, i -< j < n, such that every element in M has non-zero .th 3 row.Suppose this is not the case.Then there is a collection of elements XI, We can assume without loss of generality that every non-zero element of M has non-zero first row.Let M' be the module consisting of all matrices whose first row appears as the first row of a matrix in M, and whose other entries are zero.Define a map f: M + M' as follows: If A M, then f (A) is the matrix whose first row is the same as that of A, and whose other entries are zero.This map is certainly an R n isomorphism and M' is of the appropriate form This completes the proof.
where is defined as in Lemma 2.5.By Cor.2.4 (2), F(R) is H a two-sided ideal of R. Let X K(Rn), x the (i,j) entry of X.Then x Ei J R n Eii X Ejj K(Rn) so that H(i) x Eij 0. As in the proof of Lemma 25, this shows R that x annihilates the critical R-module .Since M is an arbitrary cyclic Rn module, R The reverse so is ; thus, by Cor 2.4 (1), x K(R) Therefore, K(R n) (K(R)) n inclusion follows by reversing the steps of the argument.Hence K(R n) (K(R)) n when R has identity If R does not contain identity, embed R into R I. From the previous paragraph and Thin.2.2 we have (K(R)) n (K(RI))n K((RI)n).However, just as a critical module over R can be considered a critical module over R 1 and vice versa, so we can identify modules over Rn and (R l)n.Therefore, by Cor. 2.4 (i), K((RI)n) K(Rn).This completes the proof.
We now describe the relation between the Krull radical of a ring R and that of a two-sided ideal I in R.
LEMMA 2.8.Let R be a ring such that R K(R), let I be a two-sided ideal of R, and let M be a k-critical 1-module.Then either MI 0 or MI is a k-critical Rmodule.

PROOF
Assume MI + 0, and take C to be a critical R-submodule of MI..4 (1), so CI 0. Hence K dim C R K dim C I k, which implies that K dim MI R > k.Since the reverse inclusion always holds, we have K dim MI R k.That MI is a critical R-module follows from the fact that MI is a critical 1-module.
Let R be a ring such that R K(R), and let I be a two-sided ideal of R. Then K(1) I.

PROOF
Let M be a critical right 1-module.If MI + 0, then there is some i I for which Mi + 0. Since MiR _ MIR 0 by Lemma 2.8 and Cor. 2.4 (i), Mi is a critical R-module.Hence the map f: M -Mi defined by f(m) mi for all m M is actually an 1-isomorphism.But then for any m M, f(mi) f(m)i mi 2 0 and hence Mi 0, contradiction.Therefore MI 0, and by Cor. 2.4 (i), I K(1).EXAMPLE 2.10.Prop. 2.9 is true if we substitute the Jacobson radical for the Krull radical.By [4, T. 48], this is equivalent to the fact that J(1) I n J(R) for any ideal I of a ring R. Unfortunately, this does not hold for the Krull radical.

F F[ x]]
Let R where F is any field, x is a couting indeteinate over F, 0 F[x]J and the ring operations are the usual matrix addition and multiplication.By [i, 0 gx. 4 K(R) 0. However, if we take I then K(I) I because I has 0 0 no special co-critical right ideals; for, since I I is isomorphic to a direct sum of copies of F, any special co-critical right ideal would have to be maximal modular.
Certainly I has no such right ideal.
We now describe the containment relations between K(R) on the one hand and J(R) and P(R) (the prime radical of R) on the other.
(i) For any ling R, K(R) S J(R).
(2) If R is a ring with Krull dimension, then K(R) __c P(R).
(3) If R is a commutative ring, then P(R) c K(R).

PROOF
If we embed R into RI, then P(R) P(RI), J(R) J(RI), and.K(R) K(R I) by [4, Cot. after Thm.59], [3, p. ii Thm.2], and Thm. 2.3 of this paper respective- ly.Hence we may assume that R has identity.Now (i) follows from the definitions of K(R) and J(R), while (2) and (3) are mentioned in [i, p. 188] for rings with identity.

Z2[Xl, x2, Xn
] where {xl, x 2 x n, } is a countably infinite set of commuting indeterminates.Take I to be the ideal generated by the polynomials x2j_l x2j + S x2j+l x2j+2, j i, 2, and let R .Say that x2j_l x2j x in R for all J.
(3) In general, P(R) and K(R) are incomparable.Let I where these symbols are defined in the previous paragraph.Then a e P(R), but a K(R); for, if I' is the

Z2 (I' + I) ]
ideal of S generated by all x i, j > i, then C is critical but 0 0 S C a + 0. Now b P(R), but b K(R), because a map f: / M, where M has Krull dimension, has kernel containing almost all the x.'s, and hence x.

CO-PRIMITIVE IDEALS
Just as J(R) can be expressed as the intersection of certain two-sided ideals of R, so can K(R).Let HI, H 2, Hn be a finite collection of special co-critical right ideals of RI, and suppose that E() E() for all 1 < J, k < n.If 3 n R K dim ?. k for all i < j _< n, then the largest two-slded ideal D c__ o Hj is called 3 j=l a k-co__-primltive ideal of R.An ideal which is k-co-primitive for some ordinal k is R I r 0 for all called co-primitive.It fs not hard to see that D {r R Hj) I i < j -< n}.Here (H')I3 is the extension of H. to a co-critical right ideal of R I as in Lemma 2.1.R, THEORY4 3.1.
K(R) is the intersection of all the co-prlmitive right ideals of PROOF From Cot. 2.4 (2), K(R) is a two-sided ideal of R. Since K(R) H for every special co-critical right ideal H _ R, then K(R) _ D for every co-primitive ideal of R. Thus, K(R) n D. Conversely, if r 0 D, then by the observation previous to this theorem we have M r 0 for any critical R-module M. Thus, If 0 is a k-co-primitive ideal of a ring R with Krull dimension k, then R is said to be k_-primitive.This definition coincides with that given in [6].
PROPOSITION 3.2.Let R be a ring with Krull dimension k.Then R is k-primitive if and only if R has a faithful critical finitely generated module C with K dim R K dim C. PROOF Suppose that R is k-primitive.Then there is a finite collection of special k-co-critical right ideals HI, Hn whose intersection is 0 and such that E E for all I j, k n.But then E((Hj)I E( so we may assume that each (Hj)' lles in the same injective hull.The module C (HI) + + (Hn) I is critical by [6, Lemma 3.1], finitely generated, and faithful, and K dim R k K dim C. The converse follows by reversing the steps of this argument.
The main properties of k-primitive rings have been investigated in [6].We list some of these properties here.Recall that the assassinator of a uniform module C over a ring R with Krull dimension is that ideal P which is maximal among the annihilators of submodules of C. THEOREM 3.3.Let R be a k-primitive ring with faithful critical module C, and R. TUCCI let P be the assassinator of C.
(1) If A, B 0 for two right ideals A and B, then either A 0 or B P (i.e., R is P-primary) (2) P is the only prime ideal of R which is not a large right ideal; (3) if H is any non-zero right ideal of R, then K dim H K dim R; (4) R and C are nonsingular; R (5) if H is a large right ideal of R, then K dim < K dim R; (6) the injective hull of R is a simple artinian ring.
In [7, Thm.3.4], Boyle, Deshpande and Feller characterize a k-primltive piece- wise domain (PWD) which contains a faithful critical right ideal.(We shall refer to this type of ring as a BDF ring after the authors.)This result can be used to de- scribe a slightly broader class of rings.Recall that a PWD R is a ring with identity which contains a complete set of orthogonal idempotents e l, en such that if x e i R ej, y E ej R ek, then x y 0 implies x 0 or y O.In what follows, we assume that R is written as an n x n upper triangular matrix ring; see [8].Recall also that a ring S is a quotient ring of R if R is a large R-submodule of S.
In the next result, we assume that R is a noetherian k-primitive ring with identity which is a direct sum .ofnon-isomorphic critical right ideals (and hence is a PWD by [8]).Since E(R) is a matrix ring over a division ring D with identity I, we can define the matrix M Ell + + Eln where Eli is the matrix with i in the (l,j) position and O's elsewhere, i < j < n.PROPOSITION 3.4 Let R and M be as above.Then R has a quotient ring S R + RMR 2 which is a noetherian BDF ring if and only if (RMR) _ RMR + R and RMR is a finitely generated R-module.

PROOF
Note that R is an upper triangular matrix ring with ejRe k 0 for j > k.
Also, each e Re. is noetherian, for if I E ekR + e Re then e.Re R J J kj >j j 3 J Y" Finally, note that R elS.
Let S R + RMR.Assume that RMR is a finitely generated R module and that 2 (RMR) R + RMR.Since S E(R), S is a quotient ring of R. Also, S is a finitely generated R-module, which implies that S is a noetherian ring and that elS is a finitely generated R-module.Now elS _c E(elR which is uniform, so elS is a critical R-module by [9,Cor. 2.4].Let 0 H be an S-submodule of elS.Then elS elS Further, n n is merely a sum of copies (elSen) (elSen)e Re (elSen) This together with (.i) shows that elS is a critical S-module.Now elS is faithful; for, if elS s 0 for some s S, then for any idempotents ej, e k R we have elSej ejsek 0. Since S is a PWD, ej se k 0. Therefore, s 0.
Conversely, let S R + RMR be a noetherian BDF ring.Since S S is noetherian (elSen) is noetherian.But by (3.2), (elSen) is noetherian.Let S'R, so that (elSen_ I + elSe n) is noetherian.Continuing en-iRen-i R in this manner, we have elSR noetherian, and hence R is finitely generated.2 Finally, since S is a ring, (RMR) __c R + RMR..4 applies more generally to a ring R with identity which is a direct sum of non-singular non-isomorphic critical right ideals; such a ring is a direct sum of ideals, each of which is a k-primitive ring by [i0, Prop.(2) Let x and y be commuting indeterminates over F, and let has no Krull dimension, since Fix, y] does not have finite uniform dimension as an F[y] module.In this case RMR is not finitely generated.

CRITICAL RINGS
A ring R is critical if is a critical R-module.If R has identity, then is faithful, and hence R is k-primitive.Thus, we could describe the structure of this ring using Thm. 3.3.However, it is possible to prove more about R, even if we do not assume that R has identity.
PROPOSITION 4.1.If R is a domain with Krull dimension, then R is critical.

PROOF
Let C be a critical right ideal of R, 0 c E C. The map f: R C given by f(r) cr is i-i, proving R is critical.
The converse of Prop. 4.1 is true if R has identity.To examine this converse for k-critical rings which do not possess identity, we need to consider separately the cases k > 1 and k i. Recall that a module M is monoform if, for any submodule N M, a homomorphism f: N M is either zero or i-I.Any critical module is mono- form by [2, Cor.PROPOSITION 4.2.If R is a k-critlcal ring, k > i, then R is a domain.
k th row of k is zero.But then n n (AXI)R n n(AXn)R n 0 contradicting the fact that M is a uniform module by [2 n Cor.2.5 and 2.6]. 5.3].

EXAMPLE 3 . 5 .
(i) Let F be a field, x a commuting indeterminate over F, 2.5].