In this lecture I want to present a couple of short results that are nevertheless very useful in practice when trying to show that a given polynomial in is irreducible. Of course, we may assume that the polynomial actually has integer coefficients. In this case, it turns out that analyzing whether the polynomial factors over suffices.
Lemma 1 () Let
- Suppose that with Let be a prime number. If divides all the coefficients of then either it divides all the coefficients of or else it divides all the coefficients of
- Suppose that with Then there is a rational such that, letting and then In other words, if is irreducible over then it is irreducible over
Proof: To fix notation, let’s say that
1. Suppose first (i.e., all the and all the are integers), and let be a prime that divides all the coefficients of their product. Towards a contradiction, suppose that does not divide all the coefficients of and does not divide all the coefficients of By the well-ordering principle there is then a least such that
and, similarly, there is a least such that
Now compare the coefficient of in both sides of the equation
where, if necessary, we set if and if
by minimality of Similarly,
by minimality of
We are assuming that so we are forced to conclude that This is a contradiction, since is prime but and
2. Now suppose that that and that By clearing up denominators, we may find a nonzero integer such that the following principle holds: for some with a rational multiple of and a rational multiple of as we can take and for appropriate integers with By the well-ordering principle, there is a least positive integer such that holds with in the place of We are done if we show that Suppose otherwise. Then there is a prime number such that Then divides all the coefficients of Then, by item 1, divides all the coefficients of or it divides all the coefficients of By dividing both sides of the equation by we find that holds with in place of This contradicts the minimality of
We can then take to obtain
The prime divides all the coefficients of so it must divide all the coefficients of one of the factors in the right hand side. A summary inspection reveals that, indeed, 2 divides the coefficients of Dividing both sides by 2 gives us the equation
The prime divides all the coefficients of so it must divide all the coefficients of one of the factors in the right hand side, in this case, Dividing by 3 shows that
As another example, consider If this polynomial factors over one of its factors must be linear, and monic, say We also must have so Since and are not roots of we conclude that is irreducible over It follows that is irreducible over
‘ lemma provides us with the following criterion that allows us to show in many cases irreducibility over
Theorem 2 (Eisenstein) Suppose that If there is a prime number such that:
then is irreducible over
Proof: By ‘ lemma it suffices to show that is irreducible over Suppose otherwise, and let
be polynomials in with and Since then and It follows that there are least integers with such that and with such that Since then either or By exchanging the roles of we may assume that so in the paragraph above. On the other hand, so so that as above is positive.
Now compare the coefficient of on both sides of the equation We have
Note that because Note also that
by minimality of
It follows that which is a contradiction, since is prime, and
For example, is irreducible over as witnessed by the prime
As a more significant example, suppose that is prime, and let’s check that
is irreducible over This does not follow directly from Eisenstein’s criterion, since no prime divides 1.
We know that Let so and
for all with since
is an integer and is a factor of the numerator but not of the denominator.
Since and Eisenstein’s criterion shows that is irreducible over and therefore so is
Corollary 3 If in Theorem 2 the conditions on are instead:
then again irreducible over
Proof: Let Then Eisenstein’s theorem applies to
For example, if then
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