Fakultät für Mathematik Sommersemester 2017 JProf. Dr. Christian Lehn Dr. Alberto Castaño Domínguez Algebra Übungsblatt 8 (Lösungen) Aufgabe 1. Es seien R ein Ring, m R ein maximales Ideal und M ein R-Modul. Wir haben gesehen, dass die R-Modulstruktur auf M/mM diesen Modul zu einem Vektorraum über k = R/m macht. Zeigen Sie, dass M/mM endliche Dimension über k hat, falls M endlich erzeugt als R-Modul ist. Zeigen Sie, dass folgende Umkehrung gilt, wenn (R, m) ein lokaler Ring 1 ist: ist m 1,..., m n eine k-basis von M/mM, so erzeugen m 1,..., m n den Modul M. Hinweis: Es sei N M der von allen m i erzeugte Untermodul. Wenden Sie Nakayamas Lemma auf M/N an. Beweis. The first part is easy: If there exists some n N such that M is generated, say by m 1,..., m n, that means that every element m M can be written as a linear combination m = a i m i, where the a i are in R. But then, taking classes modulo the submodule mm, we get that every vector m M/mM can be written as m = a i m i, where now the a i can be chosen in R/m = k and the m i in M/mM. So we have found a finite system of generators, from which we can take a finite basis, that is to say, M/mM is a finite dimensional k-vector space. The other point is a bit more subtle. What we know now is the opposite: there exists a basis {m 1,..., m n } of the k-vector space M/mM. Then for every m M, its equivalence class modulo mm can be written as m = a i m i, 1 d.h. R hat genau ein maximales Ideal.
with some coefficients a i k. Equivalently, m i a im i mm. Take N to be the R- module generated by the m i, with i = 1,..., n. Then M = mm + N, since the inclusion towards the left is trivial and the other one is shown by what we have done above. In that case, the quotient module M/N is M/N = (mm + N)/N = m (M/N), where the second equality is just the definition of quotient module when we are not sure that the denominator is included in the numerator 2. Now note that the Jacobson radical of R is m, because of it being a local ring, so by Nakayama s Lemma, M/N = 0, that is to say, M = N = R m 1,..., m n. Aufgabe 2. Man zeige, dass Q ein torsionsfreier Z-Modul ist, der nicht frei ist. Genauer: Man zeige, dass der von je zwei Elementen x, y Q erzeugte Untermodul Z x, y Q nicht frei sein kann. Beweis. It is clear that Q is torsion free over Z; if we had some a Z Q and q Q such that aq = 0, since Q is a field then either a or q vanish. Now take any two nonzero elements x, y Q. Write x = a/b, y = c/d. Then bcx ady = 0, and none of bc and ad can vanish, so we have found a non-trivial Z-linear relation for any two nonzero elements of Q, so there cannot be a Z-basis of Q. The only possibility left is that Q is generated by a single element, but that is impossible either, because for any q Q, q/2 / Z q. In conclusion, Q is a torsion free Z-module which is not free. Aufgabe 3. Es seien k R Ringe. Wir definieren den R-Modul Ω R/k als den Faktormodul E/B, wobei E := R dx x R und B den von den Elementen der Form d(x + y) dx dy, d(xy) x dy y dx, dλ, x, y R x, y R λ k erzeugten Untermodul von E bezeichnet. Der Modul Ω R/k heißt Modul der Kähler Differentiale. Wir definieren die k-lineare Abbildung d : R Ω R/k, x dx Er spielt in der Geometrie eine große Rolle. Wir definieren ferner den Modul der algebraischen p-formen von R über K als Ω p R/k := Λp Ω R/k. Verifizieren Sie die folgenden Aussagen. 2 It has no sense writing m(m/n) = mm/n if N mm, and on the other hand, if N mm obviously mm + N = mm, so there is no harm.
1. Die Abbildung d ist eine k-derivation, d.h., d ist k-linear und erfüllt die Leibniz Regel: d(xy) = xdy + ydx. 2. Das Paar (Ω R/k, d) hat die folgende universelle Eigenschaft: Für jeden R-Modul M zusammen mit einer k-derivation δ : R M gibt es genau eine R-lineare Abbildung f : Ω R/k M so, dass f d = δ. 3. Ist R = k[x 1,..., x n ], so ist Ω R/k der freie Modul, der von den dx 1,..., dx n erzeugt wird. Hinweis: Zeigen Sie zunächst, dass die dx i den Modul erzeugen. Sodann betrachte man die Derivation x i : R R und bediene sich der universellen Eigenschaft des Moduls der Kähler Differentiale. 4. Es sei wieder R = k[x 1,..., x n ]. Die Moduln Ω p R/k sind frei vom Rang ( n p) für p n und 0 für p > n. Beweis. Although Ω R/k is defined as a quotient module, we will omit any mention to B and call in the same way the elements dx of E and their equivalence classes, following the usual notation. 1. All the properties follow from the relations between the generators of Ω R/k. For instance, B contains all elements of the form d(x + y) dx dy for any pair of elements of R x and y. Then in the quotient d(x + y) dx dy = 0, so in other words, d(x + y) = dx + dy. The Leibniz rule follows from the second family of relations: since d(xy) xdy ydx = 0 in Ω R/k, then d(xy) = xdy ydx. To complete the proof of the linearity, note that d(λx) = dλx + λdx by the Leibniz rule, but we have that dλ = 0, according to the third family of elements of B. Then, d(λx) = λdx, so d is k-linear. 2. Take f : Ω R/k M defined by f(dx) = δ(x). In this way it induces automatically an R-module homomorphism, because we have determined the image of all the generators of Ω R/k 3, and by construction, it obviously holds that f d = δ. On the other hand, assume there were two such R-module homomorphisms f and g. Then, since f d = g d = δ, f and g would coincide on every element of the form dx for x R. But those are exactly the generators of Ω R/k. Again, if two R- module homomorphisms coincide on the generators of an R-module, they are equal (if something is zero on the generators, it is zero everywhere). 3 There are two ways of defining a homomorphism: either defining it as a mapping and proving it is indeed a homomorphism, or defining it (without contradictions) on the generators of our starting set (group, ring, module, vector space, etc.) and imposing it is a homomorphism.
3. By linearity, we can first claim that Ω k[x]/k is generated by all the differentials of the monomials d (x u ) = d (x u 1 1... x un n ), for u N n. Indeed, any polynomial is a k-linear combination of monomials, so its image by d will be the analogous k-linear combination of the d (x u ). On the other hand, the Leibniz rule can be generalized to an arbitrary number r of factors, in such a way that for any a 1,..., a m R, we have that d(a 1... a r ) = r a 1... â i... a r da i, where the hat over some element means that we are omitting that element from the product. Let us prove that by induction. Indeed, when r = 2 we have the original formula, so let us assume that the following holds for r = k 1. Now if r = k, d(x 1... x k ) = d ((x 1... x k 1 )x k ) = x k r x 1... ˆx i... x r dx i + x 1... x k 1 dx k, which is what we were looking for. Then we can apply this generalized Leibniz rule to our product x u = x 1 (u.. 1). x 1... x n (un)... x n, to get that d (x u ) = d u i x u e i dx i, where e i is the n-tuple having all zeros except one 1 at the i-th coordinate. All the discussion above shows that in fact every differential d(p) with p k[x 1,..., x n ] can be written as a R-linear combination of the dx i. Now the question is whether this generating system is also linearly independent or not. As indicated in the hint, consider the derivation i := / x i : k[x] k[x]. By the universal property of the module of Kähler differentials, there must be a homomorphism of R-modules f i : Ω k[x]/k k[x] extending i in such a way that f i d = i. Assume now that we had a vanishing linear combination 0 = n j=1 p jdx j Ω k[x]/k. Then, by linearity and the paragraph above, 0 = f i ( j=1 p j dx j ) = p j (f i d)(x j ) = j=1 p j i (x j ) = p i. j=1 Since the calculation did not depend on the value of i, we can deduce that for every i = 1,..., n, p i = 0, so the generating system {dx 1,..., dx n } is also a basis. 4. Now we know that Ω p k[x]/k = Λp Ω k[x]/k. In fact, we could forget about the particular expression for our module of differentials and only focus in that it is a free object of
rank n. Let us call it Ω to simplify a bit the notation. Λ p Ω is a quotient of the tensor product T p Ω := Ω. (p).. Ω, so its elements are of the form a 1... a p. Paying attention to the submodule of T p Ω with which we take quotients (seen in the lecture) we can see that the wedge product is R-multilinear and antisymmetric, that is to say, (a + b) c = a c + b c, a (b + c) = a b + a c, ab c = b ac = a(b c) and b c = c b, for a, b, c Ω 4. Since it preserves R-linear combinations, every element of Λ p Ω will be as well an R-linear combination of elements of the form dx i1... dx ip, for i 1,..., i p {1,..., n}. This would give us at most n p generators. However, by the antisymmetry, in the expression above i j and i k must differ for every j k, since a a = a a and so a a = 0. So we only need to take as a generating system the set of all the dx i1... dx ip, where the i j are all pairwise distinct (we can assume moreover that i 1 < i 2 <... < i p ). This gives us an amount of ( n p) if p n, and if p > n, all the wedge products dx i1... dx ip must vanish by the pigeonhole principle (there are n variables and p > n differentials, so there must be at least two equal dx i ). Why those systems are also basis? We will prove it by induction on n p, using as well a simple fact: by definition of Λ p Ω, if adx i1... dx p = 0, then a = 0 for any p. If n = p, then it is easy; we have a unique element in the generating system which cannot be zero by construction. Assume the result is true for p = n k and let us show it for p = n k 1. Consider a vanishing linear combination r a i dx i,j1... dx i,jp. Unless we have only one summand, in which case the unique coefficient must be zero by the observation above, there will be some index k {1,..., n} such that dx k does not appear at every summand of the combination, or in other words, such that x k x i,jl for some i and every l = 1,..., p. Then, by taking wedge products with dx k we will get a vanishing linear combination of at most r differential forms in Λ p+1 Ω, so by the induction hypothesis all the coefficients of that new combination must be zero. So in our original combination we have now less coefficients. Apply the same argument until you have a unique summand, and we are done. Aufgabe 4. Es seien R ein Ring und 0 M M M 0 eine exakte Sequenz von R-Moduln. Wir wollen prüfen, wann für einen R-Modul N die Sequenz 0 M R N M R N M R N 0 wieder exakt ist. Überprüfen Sie dies in den folgenden Beispielen. 4 If you do not see why, remember what we said in the first point of this exercise.
1. 0 Z n Z Z/nZ 0, R = Z und N = Z/nZ. x 2. 0 k[x] k[x] k 0, R = k[x] und N = k, wobei Skalarmultiplikation auf N mit einem Polynom p k[x] durch p.λ := p(1) λ gegeben sei. Beweis. The first question is how to pass from a homomorphism of R-modules f : M P to another f id N : M R N P R N. It can be done using the universal property of the tensor product. Namely, consider the mapping f id N : M N P R N (m, n) f(m) n. It is R-bilinear because f is R-linear 5, so it gives rise to a unique homomorphism of R- modules f id : M R N P R N such that over the generators of the tensor product we have f id(m n) = f(m) n. Now it is clear that we can consider the sequence of the statement 0 M R N M R N M R N 0. Whether it is exact or not, we have to see it with each concrete sequence. 1. So let us begin with the first one. By the discussion above, using the same notation, the homomorphism f id now acts as (f id)(x y) = nx y. But pay attention to the structure of Z-module of the tensor product Z Z Z/nZ: nx y = n(x y) = x (n y) = x ny = x 0 = 0, so f id sends all the generators of Z Z Z/nZ to zero, and thus it is the zero homomorphism, which is obviously not injective. So the sequence is not exact, at least at the first place. One could wonder whether it is still exact at the other two places, and the answer is yes. Note that now, since f id = 0, proving that the rest of the sequence is exact means showing that g id is bijective. Let us do so. First of all we should know how it is defined. g was just the canonical projection Z Z/nZ, so g id will be defined by g id(x y) = x y. It is clearly surjective, because the generators of Z/nZ Z Z/nZ are all of the form x y, which are in the image of g id, so im(g id) = Z/nZ Z Z/nZ. Regarding the injectivity, dealing with the elements of a tensor product is not easy at all, because we only have an easy expression for its generators, not for all the elements. Therefore, each time we want to prove the injectivity of a homomorphism of Z-modules starting from a tensor product, it is better to construct a left inverse 6. 5 Exercise! 6 Why only a left inverse (say for instance some i : Z/nZ Z Z/nZ Z Z Z/nZ such that i(g id) = id Z Z Z/nZ) and not an usual inverse? We will see it in a while.
Let us do it in this case, but note that it has to start from another tensor product, so we have to use again the universal property of tensor products to define it properly. Consider then the Z-bilinear mapping ĩ : Z/nZ Z/nZ Z Z Z/nZ given by ĩ(x, y) = x y. It is well defined because if x = z, then x = nk + z for some integer k, so x y = nk y + z y = k ny + z y = z y. It is also clearly Z-bilinear, so it defines a homomorphism of Z-modules i : Z/nZ Z Z/nZ Z Z Z/nZ given by i(x y) = x y. Now it is easy to see that i(g id) = id Z Z Z/nZ, because they are identical over the generators of Z Z Z/nZ (as we have already seen when proving that ĩ was well-defined). Then, since i(g id) is bijective, g id is also injective and bijective in the end. 2. Here happens more or less the same as before. Now our homomorphism f id is defined by (f id)(p(x) α) = xp(x) α. But as before, we must pay attention to the structure of k[x]-module of the tensor product k[x] k[x] k: xp(x) α = x(p(x) α) = p(x) (x α) = p(x) (x(0)α) = p(x) 0 = 0. As before, f id is the zero homomorphism, which is not injective. Let us act as in the previous point and see that the rest keeps on being exact. Remember that if our first homomorphism is zero, then saying that the sequence is exact is equivalent to say that the second one is an isomorphism. In this case g was given by p(x) p(0), so g id will be defined by (g id)(p(x) α) = p(0) α. It is again easy to see that it is surjective, because it reaches all the generators of k k[x] k, so im(g id) = k k[x] k. The discussion above about the injectivity is always valid (and convenient to keep in mind), so let us find better a left inverse. Let then ĩ be the k[x]-bilinear mapping ĩ : k k k[x] k[x] k defined by ĩ(α, β) = α β, where α is now seen as a constant polynomial. Let us see that it is k[x]-bilinear. It clearly preserves the sums, and regarding the product by a polynomial, ĩ(p(x) α, β) = ĩ(p(0)α, β) = (p(0)α) β = p(0)(α β) = α (p(0)β) = = α (p(x) β) = p(x)(α β) = p(x)ĩ(α, β), where the first, fourth and fifth equalities follow from the action of k[x] on k and the third from the action of k[x] on k[x] (we treat there p(0) as a constant polynomial). If we multiply by p(x) the second component, then we can begin the chain of equalitites at α p(0)β and the result is still valid. Then ĩ defines a homomorphism of k[x]-modules i : k k[x] k k[x] k[x] k given by i(α β) = α β. Now it is easy to see that i(g id) = id k[x] k[x] k: for any generator p(x) α, we have that i(g id)(p(x) α) = i(p(0) α) = p(0) α = p(0)(1 α) = 1 (p(0)β) =
= 1 (p(x) β) = p(x)(1 β) = (p(x) 1) β = p(x) β, following the same argument as in the previous paragraph. injective and bijective as well. Then g id is also In general, for any exact sequence 0 M M M 0 we always have an exact sequence of the form M R N M R N M R N 0. Whether we can add a zero at the left or not, being then f id injective or not, depends on a property of N as an R-module, called flatness ( Flachheit in German). As always, that is another story...