18.781 Problem Set 10 solution supplement
comments on problem 2
2. Suppose that α is an algebraic integer, and that
f (x) = xm + am−1 xm−1 + · · · + a1 x + a0
is the monic polynomial of lowest degree satisfied by x. Prove that α−1
is an algebraic integer if and only if a0 = ±1.
This problem depends on understanding some “high school algebra” facts about
factoring polynomials with integer coefficients. I put “high school algebra” in quotes
because I expect very few of you learned these facts in high school. Remember that
if R is any ring, then R[x] is the ring of polynomials with coefficients in R. Therefore
Z[x] means polynomials with integer coefficients, and Q[x] means polynomials with
rational coefficients. Recall also that the degree of a non-zero polynomial is the
highest power of x having a non-zero coefficient. The rough version is this: if a
polynomial in Z[x] can be factored in Q[x], then it can be factored in Z[x].
Definition. Suppose F is a non-zero polynomial in Z[x]. The content of F ,
written c(F ), is the greatest common divisor of the coefficients of F ; it is a positive
integer. We say that F is primitive if its content is 1; that is, if the only common
factor of the coefficients is 1.
Clearly we can write F = c(F ) · F0 , with F0 a primitive polynomial (with integer
coefficients).
Definition. Suppose G is a non-zero polynomial in Q[x]. Let d be the least common denominator for all the fractional coefficients of G (written in lowest terms);
this is a positive integer, and F = dG has integer coefficients. Write c = c(F ) for
the content of F , so that F = cF0 with F0 primitive, and
G=
c
F0 .
d
The content of G, written c(G), is the positive rational number c/d.
In this definition, no prime divisor p of c can divide d; for that would mean
that every coefficient of G was written with a factor of p in both numerator and
denominator. So c/d is in lowest terms. It is easy to check that the expression of
G as a positive rational number times a primitive polynomial is unique.
Lemma. Suppose that F0 and G0 are primitive polynomials in Z[x].
Then H = F0 G0 is primitive as well.
Proof. Suppose not. Then the content of H is greater than 1; so there is a prime
number p dividing c(H). This means that p divides every coefficient of H. That
is, the reduction of H modulo p is zero. The factorization of H can be reduced
mod p as well: H (mod p) = [F0 (mod p)][G0 (mod p)]. Since F0 is primitive, p
cannot divide all the coefficients of F0 ; so F0 (mod p) is a non-zero polynomial in
Z/pZ[x]. Similarly G0 (mod p) 6= 0. But the product of non-zero polynomials with
coefficients in a field must be non-zero, so it follows that H (mod p) cannot be
zero. This is a contradiction, proving the lemma. Q.E.D.
We now return to the world of Problem 2. If α is zero, then the monic polynomial
of lowest degree satisfied by α is x; so a0 = 0 and α−1 is not an algebraic integer, as
2
the problem claims. So we may as well assume that α 6= 0. If a0 were zero, then α
would also satisfy the monic polynomial f /x, which has degree m − 1, contradicting
the choice of f ; so a0 6= 0. I claim that
the polynomial f cannot be factored in Q[x].
Suppose that it can; then
f = G 1 · G2 ,
with G1 and G2 rational polynomials of degrees m1 and m2 , with m1 and m2
positive integers and m1 + m2 = m. Write
G1 =
c1
f1 ,
d1
G2 =
c2
f2 ,
d2
with f1 and f2 primitive with integer coefficients. Our factorization of f now reads
f=
c1 c2
f1 f2 .
d1 d2
Since f is monic with integer coefficients, it is primitive; and f1 f2 is primitive by
the lemma. By the uniqueness mentioned after the definition of content, we must
have c1 c2 /d1 d2 = 1; that is, f = f1 f2 , a factorization by integer polynomials. Since
f has leading coefficient 1, the fi must have leading coefficient ±1; so after changing
signs, we may arrange for fi to be monic. Since f (α) = 0, one of the factors fi (α)
must be zero. This contradicts the choice of f to have lowest possible degree, and
so proves that f cannot be factored.
Now it is fairly easy to prove that
rational polys G with G(α) = 0 are those divisible by f in Q[x].
Similarly,
integer polys F with F (α) = 0 are those divisible by f in Z[x].
Suppose now that α−1 is an algebraic integer. Write h for the monic integer
polynomial of smallest degree m0 satisfied by α−1 . Write
g(x) = a0 xm + · · · + am−1 x + 1,
a primitive polynomial of degree m. (It’s primitive because one of its coefficients is
1.) The posted solutions explain that g(α−1 ) = 0. By the second fact stated above
for α−1 , g must be divisible by h; so m ≥ m0 . Reversing the coefficients of h gives
a polynomial satisfied by α, so in the same way m0 ≥ m. Therefore m0 = m, so
g = ch for some integer c. Since g h are both primitive, c = ±1; examining leading
coefficients shows that a0 = ±1, as we wished to show.