SPLITTING OF SHORT EXACT SEQUENCES FOR MODULES. 1. Introduction Let R be a commutative ring. A sequence of R-modules and R-linear maps.

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1 SPLITTING OF SHORT EXACT SEQUENCES FOR MODULES KEITH CONRAD 1. Introduction Let R be a commutative rin. A sequence o R-modules and R-linear maps N M P is called exact at M i im = ker. For example, to say M h P is exact at M means h is injective, and to say N h M is exact at M means h is surjective. The linear maps comin out o or oin to are unique, so there is no need to label them. A short exact sequence o R-modules is a sequence o R-modules and R-linear maps (1.1) N M P which is exact at N, M, and P. That means is injective, is surjective, and im = ker. Example 1.1. For an R-module M and submodule N, there is a short exact sequence N M M/N, where the map N M is the inclusion and the map M M/N is reduction modulo N. Example 1.2. For R-modules N and P, the direct sum N P its into the short exact sequence N N P P, where the map N N P is the standard embeddin n (n, ) and the map N P P is the standard projection (n, p) p. Example 1.3. Let I and J be eals in R such that I + J = R. Then there is a short exact sequence I J I J + R, where the map I J R is addition, whose kernel is {(x, x) : x I J}, and the map I J I J is x (x, x). This is not the short exact sequence I I J J as in Example 1.2, even thouh the mdle modules in both are I J. Any short exact sequence that looks like the short exact sequence o a direct sum in Example 1.2 is called a split short exact sequence. More precisely, a short exact sequence N M P is called split when there is an R-module isomorphism θ : M N P such that the diaram (1.2) N M P θ N N P P 1

2 2 KEITH CONRAD commutes, where the bottom maps to and rom the direct sum are the standard embeddin and projection. In Section 2 we will ive two ways to characterize when a short exact sequence o R- modules splits. Section 3 will discuss a ew consequences. Beore doin that, we want to stress that bein split is not just sayin there is an isomorphism M N P o R- modules, but how the isomorphism works with the maps and in the exact sequence: commutativity o (1.2) says : N M behaves like the standard embeddin N N P and : M P behaves like the standard projection N P P. (Notice that the outer vertical maps in (1.2) are both the entities.) Example 1.4. Let R = Z, so R-modules are just abelian roups. Fix a positive inteer a > 1. Set N = Z, P = (Z/aZ) N (a countable direct sum o copies o Z/aZ indexed by {, 1, 2,...}), M = N P. Even thouh M literally equals the direct sum o N and P, we will build a short exact sequence (1.1) with a nonstandard injection : N M and surjection : M P so that (1.1) is not split. Deine : N M by (x) = (ax, ), and : M P by (x, y, y 1,...) = (x, y, y 1,...). In words, the map reduces the irst component modulo a and shits each o the other components over one position. Both and are Z-linear (i.e., they are additive), is obviously injective, is obviously surjective, and im = az = ker, so (1.1) with these modules M, N, P and linear maps, is a short exact sequence. We will show there is no isomorphism θ : M N P makin (1.2) commute by showin M ails to have a property relative to its submodule (N) that N P has relative to N. In a direct sum o abelian roups A B we have c(a B) A = ca or all c Z, where A is entiied with A {}. In words, this equation says an element o A that is a c-multiple in A B is a c-multiple in A. (Similarly, c(a B) B = cb.) Obviously ca c(a B) A. For the reverse containment, i c(a, b) = a, where a really means (a, ), then ca = a (and cb =, but we inore this), so c(a B) A ca. I there were an isomorphism θ : M N P makin (1.2) commute then applyin θ 1 to the equation c(n P ) N = cn would ive us cm (N) = c(n), where c is an arbitrary inteer. This last equation ails when c = a: since (x) = (ax, ) and ap =, we have am = az = (N), so am (N) = (N), which strictly contains a(n) since a > 1. The lesson o this example is that bein split is not about a module bein isomorphic to a direct sum in an arbitrary way: the maps in the short exact sequence matter just as much as the modules. Theorem 2.1. Let N M The ollowin are equivalent: 2. When a Short Exact Sequence Splits P be a short exact sequence o R-modules. (1) There is an R-linear map : M N such that ((n)) = n or all n N. (2) There is an R-linear map : P M such that ( (p)) = p or all p P. (3) The short exact sequence splits: there is an isomorphism θ : M N P such that the diaram (1.2) commutes.

3 SPLITTING OF SHORT EXACT SEQUENCES FOR MODULES 3 I we replace R-modules with roups and R-linear maps with roup homomorphisms, conditions (1) and (2) are not equivalent: or a short exact sequence 1 H G K 1, (1) corresponds to G bein a direct product o H and K while (2) corresponds to G bein a semirect product o H and K. The reason (1) and (2) are no loner equivalent or roups is related to noncommutativity. For an exact sequence o abelian roups, (1) and (2) are equivalent (this is the special case R = Z, since abelian roups are Z-modules). Proo. We will irst show (1) and (3) are equivalent, and then (2) and (3) are equivalent. (1) (3): Deine θ : M N P by θ(m) = ( (m), (m)). Since and are R-linear, θ is R-linear. To see that the diaram (1.2) commutes, oin around the top and riht o the irst square has the eect n (n) θ((n)) = ( ((n)), ((n))) = (n, ) and oin around the let and bottom has the eect n n (n, ). Goin both ways around the second square sends m M to (m) P. To see θ is injective, suppose θ(m) = (, ), so (m) = and (m) =. From exactness at M, the condition (m) = implies m = (n) or some n N. Then = (m) = ((n)) = n, so m = (n) = () =. To show θ is surjective, let (n, p) N P. Since is onto, p = (m) or some m M, so p = (m) = (m + (x)) or any x N. To have θ(m + (x)) = (n, p), we seek an x N such that n = (m + (x)) = (m) + ((x)) = (m) + x. So deine x := n (m). Then m + (x) = m + (n) ( (m)) and θ(m + (x)) = ( (m + (x)), (m + (x))) = (n, (m)) = (n, p). Thus θ is an isomorphism o R-modules. (3) (1): Suppose there is an R-module isomorphism θ : M N P makin (1.2) commute. From commutativity o the second square in (1.2), θ(m) = (, (m)). Let the irst coordinate o θ(m) be (m): θ(m) = ( (m), (m)). Then : M N. Since θ is R-linear, is R-linear. By commutativity in the irst square o (1.2), θ((n)) = (n, ) or n N, so ( ((n)), ((n))) = (n, ), so ((n)) = n or all n N. (2) (3): To et an isomorphism M N P, it is easier to o the other way. Let h: N P M by h(n, p) = (n) + (p). This is R-linear since and are R-linear. To show h is injective, i h(n, p) = then (n) + (p) =. Applyin to both ses, ((n)) + ( (p)) =, which simpliies to p =. Then = (n) + () = (n), so n = since is injective. To show h is surjective, pick m M. We want to ind n N and p P such that Applyin to both ses, we et (n) + (p) = m. ((n)) + ( (p)) = (m) p = (m).

4 4 KEITH CONRAD So we deine p := (m) and then ask i there is n N such that (n) = m ((m)). Since im = ker, whether or not there is such an n is equivalent to checkin m ((m)) ker : (m ((m))) = (m) ( ((m)) = (m) (m) =. Thus h: N P M is an isomorphism o R-modules. Let θ = h 1 be the inverse isomorphism. To show the diaram (1.2) commutes, it is equivalent to show the lipped diaram N N N P M P commutes (h = θ 1 )). For n N, oin around the irst square alon the let and top has the eect n n (n), and oin around the other way has the eect n (n, ) h(n, ) = (n) + () = (n). In the second square, or (n, p) N P oin around the let and top has the eect (n, p) (h(n, p)) = ((n)) + ( (p)) = + p = p, while oin around the other way has the eect (n, p) p p. (3) (2): Let : P M by (p) = θ 1 (, p). Since p (, p) and θ 1 are R-linear, is R-linear. For p P, the commutativity o the diaram h M P P implies commutativity o the diaram θ N P P so ( (p)) = (θ 1 (, p)) = p. M P θ 1 N P P 3. Consequences Let s take another look at the short exact sequence in Example 1.3: (3.1) I J I J + R, where I and J are eals with I + J = R and the map rom I J to I J is x (x, x). It turns out this splits: I J is isomorphic to (I J) R in a manner compatible with the maps in the short exact sequence. That is, the diaram I J I J + R θ I J (I J) R R

5 SPLITTING OF SHORT EXACT SEQUENCES FOR MODULES 5 commutes or some isomorphism θ. The bottom row is the usual short exact sequence or a direct sum o R-modules. To show the sequence (3.1) splits, we use the equivalence o (2) and (3) in Theorem 2.1. From I + J = R we have x + y = 1 or some x I and y J. Let : R I J by (r) = (rx, ry ). Then rx + ry = r, so is a riht inverse to the addition map I J R and that shows (3.1) splits. Althouh I J = (I J) R as R-modules, it need not be the case that either I or J is isomorphic to I J or R. Example 3.1. Let R = Z[ 5], I = (3, 1 + 5), and J = (3, 1 5). Then I + J contains 3 and = 2, so it contains 1 and thus I +J = R. From I +J = R, I J = IJ and IJ = 3R = R. Thereore I J = R R as R-modules. The eals I and J are not isomorphic to R as R-modules since they are nonprincipal eals: I 2 = (2 5), J 2 = (2 + 5), and neither ±(2 + 5) nor ±(2 5) are squares in Z[ 5]. Usin Theorem 2.1, we can describe when a submodule N M is a direct summand. Theorem 3.2. For a submodule N M, the ollowin conditions are equivalent: (1) N is a direct summand: M = N P or some submodule P M. (2) There is an R-linear map : M N such that (n) = n or all n N. Proo. (1) (2): Let : M N by (n + p) = n. This is well-deined rom the meanin o a direct sum decomposition, and it is R-linear. Obviously (n) = n or n N. (2) (1): There is a standard short exact sequence N M M/N. Since is a let inverse to the inclusion map N M in this short exact sequence, the equivalence o (1) and (3) in Theorem 2.1 implies there is a commutative diaram N M M/N θ N N (M/N) M/N where θ(m) = ( (m), m) is an R-module isomorphism. For n N, θ(n) = ( (n), n) = (n, ), so usin θ 1 shows M has a direct sum decomposition with N as the irst summand. Example 3.3. Usin R = Z, there is no direct sum decomposition o abelian roups Z = 2Z P or a subroup P o Z, since P would have to be isomorphic to Z/2Z and no element o Z has order 2. This ailure o (1) in Theorem 3.2 implies the ailure o (2), i.e., there is no additive map : Z 2Z such that (n) = n or all n 2Z, which can also be seen directly: rom (2) = 2 we have 2 (1) = 2, and there is no solution or (1) in 2Z. Theorem 3.4. For an injective R-linear map N M, the ollowin conditions are equivalent: (1) (N) is a direct summand o M. (2) There is an R-linear map : M N such that ((n)) = n or all n N. The proo is similar to that o Theorem 3.2.

6 6 KEITH CONRAD Example 3.5. In Example 1.4, where : Z Z (Z/aZ) N is an injective Z-linear map other than the standard one, we showed (Z), which is az, is not a direct summand o Z (Z/aZ) N (even thouh (Z) = Z and Z is a direct summand o Z (Z/aZ) N ). Thereore there is no Z-linear map : Z (Z/aZ) N Z such that ((n)) = n or all n Z. To check directly that there is no, suppose exists. The equation ((1)) = 1 says (a, ) = 1. Since (a, ) = a(1, ), we have a (1, ) = 1 in Z, which is impossible since a > 1.