JSH: Three proofs
Rupert
3-sqrt(-26), one is divisible by 7 in the ring of algebraic integers
and one is not. Consequently, he claims that the theorem that a
non-monic primitive polynomial over Z which is irreducible over Q
cannot have an algebraic integer as a root is false. Here, for his
convenience, are three proofs that he is wrong, two of which are
completely self-contained.
The numbers (3+sqrt(-26))/7 and (3-sqrt(-26))/7 are complex conjugates.
Consequently, if one of them is a root of a polynomial with real
coefficients, the other is a root of the same polynomial. In
particular, if one is a root of a monic polynomial with integer
coefficients, the other is as well. So either both of them are
algebraic integers or none of them are. Now, their sum is 6/7. If 6/7
were a root of a monic polynomial p(x) with integer coefficients of
degree n, then p(6/7) would have 7^n in the denominator and so could
not be 0, contradiction. So 6/7 is not an algebraic integer. So it is
impossible that both the numbers could be algebraic integers. So
neither of them are.
More generally, suppose that a_1, a_2, ... a_n are roots of a non-monic
primitive polynomial p(x) over Z which is irreducible over Q. I claim
that either they all are algebraic integers or none of them are. Let K
be a splitting field for p(x) over Q. That is, let K be the smallest
field containing Q and a_1, a_2, ... a_n. The Galois group of K over Q
clearly acts on the a's. In fact the action is transitive. That is,
given any a_i and a_j there is an automorphism of K which maps a_i to
a_j. The proof of this is given in Chapter 7 of Grayling's "A Course in
Galois Theory". Now, if a_i is a root of a monic polynomial with
integer coefficients, and sigma is an automorphism of K, then
sigma(a_i) is also a root of the same polynomial. So, as before, either
all the a's are algebraic integers or none of them are. Now, if we
divide p(x) by its leading coefficient, we get a monic polynomial p'(x)
irreducible over Q which has at least one coefficient that is not an
integer. The coefficients of p'(x) will be symmetric polynomials in the
a's. If all the a's are algebraic integers, then all the coefficients
of p'(x) will be algebraic integers. However, a rational number which
is an algebraic integer is an ordinary integer. Because if a rational
number r has a prime p in its denominator, and q(x) is a monic
polynomial with integer coefficients of degree n, then q(r) will have
p^n in its denominator and so will not be zero. So that means the
coefficients of p'(x) will all be integers, contradiction. So not all
the a's are algebraic integers. Therefore none of them are. This proves
that a non-monic primitive polynomial p(x) over Z which is irreducible
over Q cannot have an algebraic integer as a root.
Here is another proof. Let f and g be two primitive polynomials over Z.
Then their product is primitive. Given any prime p, we have that f and
g are not divisible by p. Take f and g modulo p, the results will not
be zero. Now the integers mod p are an integral domain, so fg modulo p
will not be zero. Hence fg is not divisible by p. The argument holds
for any prime p, so fg is primitive. More generally, let the content of
f be the greatest divisor of its coefficients. Then the content of fg
will be the content of f times the content of g. Now, suppose p(x) were
a non-monic primitive polynomial over Z that was irreducible over Q,
that had an algebraic integer a as a root. Since p(x) is irreducible
over Q, it divides every polynomial with coefficients in Q of which a
is root. (Otherwise the Euclidean algorithm would yield a
contradiction). So let q(x) be a monic polynomial with integer
coefficients of which a is a root. We have cq(x)=r(x)p(x) for some
integer c and some polynomial r(x) over Z. The content of p(x) and q(x)
is 1, so the content of r(x) must be c. Divide r(x) by c to get r'(x),
which also has coefficients in Z. Now we have q(x)=r'(x)p(x), which is
clearly impossible because q is monic and p isn't.