Algebra. Übungsblatt 10 (Lösungen)

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1 Fakultät für Mathematik Sommersemester 2017 JProf. Dr. Christian Lehn Dr. Alberto Castaño Domínguez Algebra Übungsblatt 10 (Lösungen) Aufgabe 1. Es sei k ein Körper. Man zeige, dass es in k[x] unendlich viele normierte Polynome gibt, die prim in k[x] sind. Beweis. We can follow step by step the great and old proof by Euclid of the infinitude of prime numbers, adapting it to our context 1. Take a finite amount of prime monic polynomials p 1,..., p n k[x], and consider p := p 1... p n + 1 k[x]. It is clearly a monic polynomial in k[x], and it will be either prime or not. If it is prime, then the amount of prime monic polynomials in k[x] is greater than n. Assume now that p is not prime, and take a prime factor q of it. q can be assumed to be monic; since k is a field, we can multiply by the inverse of the leading coefficient of q and take another prime factor of p which is monic. Now q cannot be any of the p i. If it were so, it would be also a divisor of p 1, and thus of p (p 1) = 1, which is impossible. So in the end we have anyway found a monic prime polynomial q not in the set {p 1,..., p n }. Since the proof was independent of n, we have proved that the amount of prime monic polynomials in k[x] is greater than any natural number, hence infinite. There are plenty of proofs of this fact, as with happens in the traditional case. Moreover, there are even constructive proofs. For instance, if k is infinite, all the polynomials x + α, for α k are irreducible, because they have degree one, and then prime, since k[x] is a unique factorization domain 2. The finite field case is more complicated, but we could try to show that over a field F q, with q a prime power (remember that all the finite fields have that form) all the polynomials p r (x) := x qr x 1 are irreducible (and then prime) in F q [x]. The case q = p, r = 1 is below. The rest await you, dear and interested reader. Aufgabe 2. Wir betrachten das Polynom f = x p x 1 F p [x]. Zeigen Sie, dass f irreduzibel ist. Betrachten Sie dazu beispielsweise den F p -Algebrenautomorphismus φ : F p [x] F p [x], x x + 1, und seine Wirkung auf den Primfaktoren von f. 1 A bit of history: Euclid did not proved his theorem by contradiction, as you can see here (Book IX, Proposition 20). 2 From now on, UFD (oder faktorieller Ring).

2 Beweis. As always, we will follow the indications given (why are they there otherwise?) and consider the subgroup generated by φ in Aut(F p [x]). Since φ p (a(x)) = a(x + p) = a(x) and φ id Fp[x], we can deduce that G := φ has order p. Now note that φ(f(x)) = f(x+1) = (x+1) p (x+1) 1 = x p +1 x 1 1 = f(x) 3, that is, f is invariant under the action of G defined by φ f = φ(f) 4. Therefore, The set of its prime factors must also be invariant under such an action. Take then any prime factor g of f, and consider the orbit G g, formed by prime factors of f. We know that the cardinality of the orbit must divide the order of the group, so we find two possibilities: either G g = 1 or G g = p. If the orbit has only one element, this means that g is also invariant under the action of G. But then, it cannot have less degree than f. Indeed, assume that g = p 1 i=0 g ix i has degree p 1. Then, p 1 p 1 φ(g) = g i (x + 1) i = i=0 The p 2-degree term of φ(g) = g would be i=0 g i i j=0 ( ) i x j. j (p 1)g p 1 + g p 2 = g p 1 + g p 2 = g p 2, so g p 1 = 0, and then g can only be of degree p 2. However, we could repeat the same argument until deg g = 1, deducing that g would be a constant, and so a unit or zero, which are both contradictions. Therefore, in the end g must have degree p, so f = g and g would be irreducible. What happens if the orbit of g has p different polynomials? Then, since the deg f = p, they must be all the possible factors of f, and furthermore, have degree one. But that means that f will have all its roots in F p. However, we know that for every α F p, α p = α, so 0 = f(α) = 1, which is again a contradiction. Consequently, this case cannot happen and we return to the previous one. Aufgabe 3. Es sei p eine Primzahl und f = x p 1 Z[x]. Wir schreiben f = (x 1) g mit g Z[x]. Zeigen Sie, dass g irreduzibel in Z[x] und Q[x] ist. Gilt das auch, wenn p nicht prim ist? Beweis. It is easy to see that g(x) = x p x + 1. Since the content of g is one (after all, all its coefficients are one), by lemma II.5.23, point 2, from the lecture, it will be irreducible in Q[x] if and only if so it is in Z[x]. It is complicated to show that g itself is irreducible, but we can use a little trick that makes our lives much simpler. Namely, we can apply the automorphism φ from exercise 2 to g. More concretely, g is reducible if and only if there exist two more polynomials r, s Q[x] such that g(x) = r(x)s(x), which is 3 Here we use what in other contexts is completely wrong: that (a + b) p = a p + b p for every a, b F p. 4 We do not need a proof of the fact that such rule is a group action, because G is a subgroup of Aut(F p [x]) acting in the natural way.

3 equivalent to g(x + 1) = r(x + 1)s(x + 1), that is, that g(x + 1) is also reducible (and the product of r(x) = r(x + 1) and s(x) = s(x + 1)). Let us do that then. We know that g = f/(x 1), so g(x + 1) = f(x + 1)/(x + 1 1) = f(x + 1)/x. In other words, g(x + 1) = (x + 1)p 1 x = 1 x ( p k=0 ( ) ) p x k 1 = k p k=1 ( ) p x k 1. k Now we can apply Eisenstein s criterion with p: every binomial number ( p k), for k = 1,..., p 1, is a multiple of p. Just note that ( ) p k = p! k!(p k)!, and no prime factor of k! or (p k)! can be p, because all of them are smaller integer numbers. On the other hand, neither p divides ( ) p p = 1 nor p 2 divides ( p 1) = p. Then, as said before, we can apply Eisenstein s criterion and deduce that g(x + 1) is irreducible in Q[x], and hence g(x) as well. When p is not prime, the argument above does not work. We can quickly find a counterexample. Namely, for p = 4 f = x 4 1 = (x 2 1)(x 2 + 1) = (x 1)(x + 1)(x 2 + 1), and so g = (x + 1)(x 2 + 1), which is obviously not irreducible. In fact, in general, for any even value of p we have f( 1) = ( 1) p 1 = 0, so x + 1 divides both f and g, making the latter reducible. Aufgabe 4. Zerlegen Sie das Polynom f = x 5 + 5x + 5 in irreduzible Faktoren in Q[x], in F 2 [x] und in F 3 [x]. Beweis. In Q[x] is simple, because we can apply Eisenstein s criterion again, this time with 5. Indeed, 5 divides all the coefficients of f but the leading one (they are 5 or 0), it does not divide 1, which is the leading coefficient, and 25 does not divide 5 either. In conclusion, Q[x] is irreducible in Q[x]. In F 2 [x] we will follow another strategy, which could seem crazy, but in this case is really useful. We will apply brute force and try with all the irreducible polynomials of less degree in F 2 [x]. What makes this not only doable but even effective is the tiny amount of irreducible polynomials in F 2 [x]. First of all, note that f = x 5 +x+1 has no roots in F 2, so if it were irreducible, it should be a product of an irreducible polynomial of degree 2 and another of degree 3. And checking whether a degree 2 or 3 polynomial is reducible is easy: they should have a root. There are four polynomials of degree 2: x 2, x 2 + 1, x 2 + x and x 2 + x + 1.

4 The first three of them are reducible, since have, respectively, 0, 1 and 0 and 1 as roots. Then the only irreducible polynomial of degree 2 in F 2 [x] is x 2 + x + 1. Regarding degree 3, there are eight polynomials, but we will omit those without constant term (Absolutglied), because they are multiples of x. Then we have 4 left: x 3 + 1, x 3 + x + 1, x 3 + x and x 3 + x 2 + x + 1. Both the first and the last ones have 1 as a root, contrarily to the other two. In conclusion, if f were irreducible, it should be a product of x 2 +x+1 with either x 3 +x+1 or x 3 + x An easy calculation shows that f = (x 2 + x + 1)(x 3 + x 2 + 1), and we are done. We will use the case of F 3 [x] to illustrate another method. Now f = x 5 x 1, and has no roots either. Therefore, if it were irreducible, it should have a degree 2 factor of the form d = x 2 + ax + b. If we divide f by d, we obtain a rest r (meaning that f = qd + r for some q) in F 3 [x] which has the expression r(x) = (a 4 + b 2 1)x + (a 3 b + ab 2 1). Then f has a degree 2 factor if and only if there exist a and b in F 3 such that r vanishes. We have then a system of two equations in two unknowns, namely { a 4 + b 2 1 = 0 a 3 b + ab 2 1 = 0. If a = 0, then in the second equation we get 1 = 0, which cannot be. Let us try then with a = ±1. Then in the first equation we obtain b 2 = 0, and if we replace b by 0 in the second equation we have again 1 = 0. Summing up, f cannot have a degree 2 factor, and then it is irreducible in F 3 [x]. Aufgabe 5. Es sei R ein faktorieller Ring und S R ein multiplikatives System. Zeigen Sie, dass auch S 1 R faktoriell ist. Beweis. A UFD is a ring R characterized by the following two properties: UFD1 R is an integral domain. UFD2 Every element in R can be decomposed as a product of a unit and prime elements of R (Prime decomposition, Zerlegung in Primelemente). Let us focus first on UFD1 and prove that S 1 R is also an integral domain. Let then a/s, b/t be two elements of S 1 R, and assume (a/s)(b/t) = ab/st = 0. Then, by definition of the localization of a ring, there must exist some u S {0} such that in R u(ab st 0) = 0.

5 Since R is an integral domain and u 0, then ab = 0 and either a = 0 or b = 0. But that means that either a/s = 0 or b/t = 0, having proved UFD1 for S 1 R. Let us show now UFD2 for S 1 R, assuming it holds in R. Take then and element x/s S 1 R. Since x R, there will be some prime elements p 1,..., p n and a unit u in R such that x = up 1... p n, so that x s = u n i=1 p i s = u s n i=1 p i 1. u/s is a unit, its inverse being u 1 s/1. Regarding the p i /1, we are going to show that either they are invertible or prime in S 1 R as well. Let then p be a prime in R and suppose that p/1 (a/t)(b/v) for some a/t, b/v S 1 R. Then there will exist some q/w S 1 R such that ab/tv = (p/1)(q/w), or equivalently, pqtv = abw in R. Since p is prime and divides the product abw, it will divide either a, b or w. If it divides a or b we are over, because then, in the first case for instance and writing a = pa, a/t = (p/1)(a /t), that is, p/1 divides a/t, too. Now if p w, say w = pw, then (p/1) is invertible, because (p/1)(w /w) = 1. Then, in the decomposition of x/s above, multiplying u/s and all the p i /1 that become units, we obtain a product of a unit in S 1 R and some primes in S 1 R, proving UFD2 and thus solving the exercise. Aufgabe 6. Zeigen Sie, dass die folgenden Polynome irreduzibel sind: 1. x 4 + 3x 3 + x 2 2x + 1 Q[x] 2. 3x x x 2 375x + 30 Q[x] 3. x 2 y + xy 2 x y + 1 Q[x, y] Entscheiden Sie, ob diese Polynome irreduzibel in R[x] bzw. R[x, y] sind, wenn R = Z oder R = R ist. Beweis. 1. Let f(x) = x 4 + 3x 3 + x 2 2x + 1. Since i(f) = 1, it will be irreducible in Q[x] if and only if so it is in Z[x]. If it were reducible in Z[x], then it would also be reducible in F p [x] for every p prime. Indeed, say f = pq in Z[x]. Then f = p q, where ḡ is the polynomial resulting after reducing its integer coefficients modulo p. Let us do as in exercise 4 in the case of F 3 [x]. The only rational roots of f could be 1 or -1, but none of them is so. This can easily be seen in our situation (why?), for another

6 systematic argument, the keyword is rational root test 5 Then if f were reducible, it would be a product of two degree 2 polynomials. Divide f by d = x 2 + ax + b, for some a, b Z, and obtain the following rest: r(x) = ( 2a 3 + 3a 2 a + 2ab 3b 2)x + ( a 2 b + 3ab + b 2 b + 1). Again, if f were reducible (over any ring R!), there should exist some values of a and b in R such that r(x) = 0. There are some threes in the expression of r, so it seems that reducing modulo 3 might be a good idea. Consider the system of equations arising from imposing r(x) = 0 in F 3 [x]: { a 3 a ab + 1 = 0 a 3 b + ab 2 1 = 0. If a = 0, both equations become 1 = 0, which is impossible. If a = 1, from the first equation we can deduce b = 1, but the second equation would not hold. And if a = 1, then we can say that b = 1 from the first equation, reaching a contradiction in the second. In conclusion, f is irreducible and so is f as well, in Z[x] and Q[x]. In R[x], it is reducible: the roots of every real polynomial are either real or complex, being pairwise conjugated. Then every real polynomial is the product of degree 1 or 2 real polynomials. Since f has degree 4, it must split in a product of such smaller ones, and then is reducible. 2. This point is much easier. f = 3x x x 2 375x + 30 is reducible in R[x] because of the same argument as right above, and it is reducible in Z[x], since 3, which is not a unit in Z[x] 6, divides every coefficient and thus is a factor (prime, in fact) of f. Last, since 3 is a unit in Q we can consider g = f/3, and apply Eisenstein s criterion to g with 5. Namely, 5 divides every coefficient of g but the first one, and 25 does not divide the independent term of g, so it must be irreducible in Q[x]. This example was short, so let us exemplify here yet another way of showing the irreducibility of a polynomial, more direct or naive, but good anyway. Take g = x 4 + 5x x 2 125x + 10 and let us see it is irreducible in Z[x]. Since g has no integer roots by the rational root test (in this case they can only be ±1, ±2, ±5 and ±10), the only possibility left is that it is the product of two monic 7 degree 2 polynomials, say x 2 + ax + b and x 2 + cx + d. If we multiply them and check every coefficient, we get the following system of equations: a + c = 5 b + ac + d = 15 ad + bc = 125 bd = 10 5 Sorry, the German link does not work here because of the umlaut; just go to there via the English version, which by the way has a proof. 6 Which are the units of a polynomial ring? 7 We can assume that in this example, but not always (see next point)..

7 From the last equation we deduce four possibilities, up to an exchange of the two factors (remember we work in Z): (b, d) {(1, 10), ( 1, 10), (2, 5), ( 2, 5)}. On the other hand, c = 5 a, so replacing everything in the second equation, we get a quadratic equation of discriminant 25 4(15 b d). The unique perfect square for the possible values for b and d is 9, occurring when b = 1 and d = 10, so we are left only with such case, in which a is either 1 or 4. Replacing everything in the third equation we get a contradiction anyway, because for those values of a, b, c and d, ad + bc is always positive. So in the end, sooner or later we are wrong and then f must be irreducible. 3. Let f = x 2 y + xy 2 x y + 1. Let us now prove the three cases at once. We will work in the polynomial ring R[x, y] = R[x][y], where R can be either Z, Q or R. Let us write f = xy 2 +(x 2 1)y (x 1). The polynomial x 1 is irreducible in R[x], which is a UFD since R is so. Then x 1 is also a prime element in R[x]. Now note that x 1 does not divide x, but does divide both x 2 1 and (x 1). Moreover, (x 1) 2 does not divide (x 1), so since f is primitive (for the leading coefficient of f does not divide in R[x] the other coefficients; note that also x is irreducible in R[x]), we can apply Eisenstein s criterion over R[x] with x to deduce that f is irreducible in R[x][y]. Now, as in the previous point, we will prove the irreducibility of f in another way. We will work again in the polynomial ring R[x, y] = R[x][y]. Since f is primitive, then if it is irreducible in R[x][y], it must be the product of two degree one polynomials, ay + b and cy + d. Let us multiply them and look at each coefficient. As in the last point, we get the following system of equations: ac = x ad + bc = x 2 1 bd = (x 1) Without loss of generality, we can deduce from the first equation that a = x and c = 1. Now in the third one we must distinguish two cases (we lost all generality choosing a and c): b = α(x 1) and d = α 1, or b = α and d = α 1 (x 1), for some unit α R. Replacing those values of a, b, c and d in the second equation we obtain the following, respectively: α 1 x + α(x 1) = x 2 1 α 1 x(x 1) + α = x 2 1, and both of them are impossible; the polynomial at the left-hand side of the first equality has degree one, whereas that at the left-hand side of the second equality has nonzero term in x. In conclusion, f is irreducible..

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