# Minimal polynomial of $x$ over $K\left(\frac{p(x)}{q(x)}\right) \subset K(x)$

Let $K$ be a field, and let's consider the field of rationals functions $K(x)$. Let $t\in K(x)$ be the rational function $\frac{P(x)}{Q(x)}$, where $P,Q$ have no common factors. I have to prove that the extension of fields $K(t) \subset K(x)$ has degree $\max(\deg P,\deg Q)$.

I have to prove that the minimal polynomial of $x$ over the field $K(t)$ is the polynomial, $P(X)-tQ(X) \in K(t)[X]$. Clearly $x$ is a root of the polynomial, so it remains to prove that it's irreducible.

First, by Gauss' lemma, the polynomial is irreducible in $K(t)[X]$ iff is irreducible in $K[t][X] = K[X][t]$.

That consideration it's from a hint of the book (Dummit and Foote, Section 13.2, Exercise 18).
I suppose that if there exists a factorization, then $P,Q$ will have common factors. But I don't know how to work here, it's very confusing. Please help me. \eqalign{ & \left( {\sum\limits_{k = 0}^n {a_k (X)t^k } } \right)\left( {\sum\limits_{j = 0}^m {b_j (X)t^j } } \right) = P(X) - tQ(X) \cr & a_k (X),b_j (X) \in K[X] \cr} I don't know if the possible factorization will be in $t$ or in $X$. I'm very confused right now. Sorry for being so stupid.

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Hint: It's a linear polynomial in $t$. When is a linear polynomial irreducible?
Linear polynomials are always irreducible. Yeah you are right, but I'm confused, I did never used the fact that $(P,Q)=1$ So maybe it's wrong – Daniel Oct 1 '12 at 13:39
Linear polynomials are always irreducible over fields. Over an arbitrary domain a linear polynomial is irreducible iff you can't factor out any non-unit. Since $K[x]$ is not a field, you need to check that this is not possible. You will need to use that $(P,Q) = 1$. – Brandon Carter Oct 1 '12 at 14:08
In a domain, an element $r$ is called irreducible if whenever $r = ab$, then either $a$ is a unit or $b$ is a unit. The definition in Dummit and Foote is given on p. 284. – Brandon Carter Oct 1 '12 at 14:21