Let be a field, and consider the field of rational functions over (that is, the field of fractions of the domain ). Let , with and such that the degree of is strictly larger than the degree of . In this exercise, we will compute the degree of over . (Note that if has degree larger than or equal to the degree of , then we can use the division algorithm to find , and then .)
- Show that is irreducible over and has as a root (in the extension ).
- Show that the degree of as a polynomial in is the maximum of the degrees of and .
- Conclude that .
(Fields of rational functions were introduced in Example 4 on page 264 of D&F.)
We claim that is indeterminate over – that is, that does not satisfy any polynomial over . Indeed, if , then we have . Let and denote the degrees of and , respectively. Since has no zero divisors, the degree of the th summand is . Suppose two summands have the same degree; then for some and , which reduces to . Since (as we assume) , we have . In this case, we can pick out the summand with the highest degree, and be guaranteed that no other summands contribute to its term of highest degree. This gives us that for the highest degree summand; by induction we have for all the coefficients , a contradiction. (In short, no two summands have the same highest degree term. Starting from the highest of the highest degree terms and working down, we show that each is 0.)
So is indeterminate over , and in fact is essentially a polynomial ring whose field of fractions is . By Gauss’ Lemma, the polynomial is irreducible over if and only if it is irreducible over . Now , and our polynomial is irreducible in this ring since it is linear in . So in fact is irreducible over , and moreover has as a root.
The degree of in is the maximum of the degrees of and because the coefficient of each term (in ) is a linear polynomial in , which is nonzero precisely when one of the corresponding terms in or is nonzero. That is, we cannot have nonzero terms in and adding to give a zero term.
To summarize, is an irreducible polynomial over with as a root, and so must be (essentially) the minimal polynomial of over . By the preceding paragraph, the degree of this extension is the larger of the degrees of and .