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Math 323. Midterm Exam. February 28, 2013. Time: 75 minutes. (1) Let f ∈ Z[x] be a monic polynomial, and suppose f (α) = 0 for some α ∈ Q. Prove that then α ∈ Z. Solution. You could just say that by Gauss’ Lemma, f factors over Q as (x − α)g(x) iff this factorization (up to units, which is ±1) holds over Z, and thus α ∈ Z. However, the intention of the question was to prove Gauss’ lemma directly for this special case. Here’s the proof (which, in fact, is basically the proof of the general Gauss’ Lemma): by Bezout’s theorem, f (x) = (x − α)g(x). Since f is monic, g has to be monic too. Suppose α = c/d with c, d ∈ Z and (c, d) = 1. Let D be the least common denominator of all the coefficients of g. Then we have: (1) dDf (x) = (dx − c)g ′ (x), where g ′ is a polynomial with integer coefficients. Now, if the gcd of all the coefficients of g ′ is some integer m > 1, we get: m|D, since g was a monic polynomial, and so the leading coefficient of g ′ is D. Then we can cancel m on the both sides of equation (1) (and replace D with D/m). Thus, we can assume that gcd(all coefficients of g ′ ) is 1. Now, let p be any prime dividing d. Reduce the equation (1) modulo ¯ − c̄)g ′ (x) in Fp [x], where¯denotes reduction modulo p. p: we get: 0 = (dx Since Fp [x] is an integral domain, this means either d¯ = c̄ = 0, or g ′ (x) is a zero polynomial. Now, both are impossible since (c, d) = 1, so they cannot be both divisible by p, and also we assumed that gcd(all coefficients of g ′ ) was 1. (2) (a) Is 13 prime in Z[i]? (If not, factor it into primes). Solution. We have 13 = (3 − 2i)(3 + 2i); both 3 − 2i and 3 + 2i are irreducible, since their norms are equal to 13, which is prime; and we know that (a) Norm is multiplicative, so if we had 3 − 2i = αβ, then N (3 − 2i) = N (α)N (β), so this forces N (α) or N (β) to be 1; and (b) Any element of norm 1 is a unit. (b) Prove that any ideal in Z[i] contains a positive integer. Solution. Let I be an ideal, and suppose a + bi ∈ I be any non-zero element. Then (a − bi)(a + bi) ∈ I, i.e. a2 + b2 ∈ I. (c) Let (3) be the ideal generated by the element 3 in Z[i]. Describe the quotient Z[i]/(3). Solution. The representatives of the cosets are: 0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i (so it is a ring of 9 elements, containing the field of 3 elements F3 = {0, 1, 2}). To understand multiplication in this ring, note that the polynomial x2 + 1 is irreducible over F3 (because no element of F3 satisfies it). Then F3 [x]/(x2 + 1) is a field of 9 elements (more about this in the next problem). We can make an isomorphism from F3 [x]/(x2 + 1) to Z[i]/(3) by sending a + bx to the coset of a + bi, where a, b ∈ {0, 1, 2}. So, the answer is: the field of 9 elements. 1 2 (3) Construct a field of 27 elements. Solution. Note: Z/3Z × Z/3Z × Z/3Z is definitely wrong! (as a ring, this product has zero divisors). Sure, as vector spaces (modules) over F3 , this and the field of 27 elements are the same, but they are not isomorphic as rings! To construct this field, we need an irreducible degree 3 polynomial over F3 . For example, we can take f (x) = x3 − x + 1. Since its degree is 3 and it has no roots (we just check that 0, 1 and 2 are not roots), it is irreducible. Then it generates a prime ideal in F3 [x]. Since F3 [x] is a Euclidean ring and therefore a PID, this ideal is also maximal. Then the quotient F3 [x]/(f (x)) is a field. This quotient has 27 elements, because the cosets have representatives of the form a0 + a1 x+ a2 x2 , with a0 , a1 , a2 ∈ F3 . √ ]. Is it true that in the polynomial ring R[x] the prime (4) Let R = Z[ −3+1 2 elements are the same as irreducible elements? Solution. We know that R is Euclidean (with respect to the usual complex norm), so R is a PID and therefore, a UFD. Then R[x] is also a UFD. In a UFD, prime elements and irreducible elements are the same. √ (5) Find an example of an element of Z[ −5] that is irreducible but not prime. (and give a complete proof √ that it has this property). Solution. Take 2 ∈ Z[ √−5]. Then√2 is not prime, because the ideal (2) √ contains the product (1 − −5)(1 + −5) = 6, and 1 ± −5 ∈ / (2). On the other hand, 2 is irreducible: suppose 2 = αβ with α, β not units; √ then N (2) = N (α)N (β), where N (a + b −5) = a2 + 5b2 is the usual complex norm. Then: 4 = N (α)N (β). Since α and β√are not units, we get N (α) = N (β) = 2. But the norm of an element of Z[ −5] cannot equal 2; a2 + 5b2 = 2 is impossible for a, b ∈ Z. (6) Is the polynomial x6 + 30x5 − 15x3 + 6x − 120 irreducible in Q[x]? Solution. Yes, by Eisenstein’s criterion with p = 3.