Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is most likely simple, but I want to ask a small question.

Let $k$ be a field, $k(x)$ the field of rational functions in $x$. Let $t\in k(x)$ be the rational function $P(x)/Q(x)$ with relatively prime polynomials $P(x),Q(x)\in k[x]$, with $Q(x)\neq 0$.

Let $P(X)-tQ(X)$ be a polynomial in the variable $X$ and coefficients in $k(t)$.

I want to show that the degree of $P(X)-tQ(X)$ as a polynomial in $x$ with coefficients in $k(t)$ is the maximum of the degrees of $P(X)$ and $Q(X)$.

If $P(X)$ and $Q(X)$ have different degrees, this is clear. Otherwise, I suppose both have degree $n$, and the leading term of $P(X)$ has form $k_1X^n$, and the leading term of $Q(X)$ has form $tk_2X^n$. I only want to show that these terms do not cancel. If they did, then $k_1/k_2=t$, so $t$ is a constant rational function. Is there a contradiction to be had, so that this contradictory case cannot actually happen?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

If $t$ is constant, then $P$ and $Q$ are proportional; in order for $P$ and $Q$ to be both proportional and relatively prime, they must both be constant themselves; that is, $\deg(P)=\deg(Q) = 0$.

Dummit and Foote do not specify what the degree of the zero polynomial is (under some conventions, the $0$ polynomial has degree $0$; under others, $-\infty$, under yet others, it has no degree). Their Proposition 7.2.4 suggests that the zero polynomial has no degree (they say, if $R$ is an integral domain, and $p(x)$, $q(x)$ are in $R[x]$, then the degree of $pq$ is the degree of $p$ plus the degree of $q$, but never specify $p$ and $q$ nonzero). So here we would need to waffle a bit with the definition of "degree" and note that if $P-tQ=0$, then $P$, $Q$, and $P-tQ$ are all constant polynomials, which "in spirit" is the same as "have the same degree".

share|improve this answer
    
I see now, thanks. –  Norbert Wiener Nov 3 '11 at 19:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.