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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?

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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".

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I see now, thanks. – Norbert Wiener Nov 3 '11 at 19:19

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