Download Solutions - UBC Math

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.