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Assume $k$ is a field and $x$ is transcendental over $k$. If $p(x)$ and $q(x)$ be relatively prime polynomials of $k[x]$ then I want to find the minimal polynomial of $x$ over $k\left(\dfrac{p(x)}{q(x)}\right)$.

If one of them be a constant then $[k(x):k(\dfrac{p}{q})]$ is degree of the other one and the minimal polynomial is $p(T)-p(x)$, if $p(x)$ is the non-constant one, $q(T)-q(x)$ if $q(x)$ is the non-constant one.

Now what I was going to do was this;

$\big(p(x),q(x)\big)=1\Longrightarrow \exists r(x),s(x)\in k[x]\text{ s.t. }r(x)p(x)+s(x)q(x)=1$

Then $r(T)\dfrac{p}{q}+s(T)=\dfrac{1}{q}$

So $[k(x):k(\dfrac{p}{q},\dfrac{1}{q})]=\max(\deg(p),\deg(q))$ but I am looking for minimal polynomial of $x$ over $k(\dfrac{p}{q})$ and $[k(x):k(\frac{p}{q})]$. I know that this extenction degree should be $\max(\deg(p),\deg(q))$ but I don't know its proof, also I want the minimal polynomial explicitly so a proof mentioning this theorem and that theorem is not what I'm looking for.

AmirHosein Sadeghimanesh
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H.W.
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  • Similar question: https://math.stackexchange.com/questions/3353809/let-x-be-transcendental-over-f-let-y-fx-gx-be-a-rational-function-pr?rq=1 – 19021605 Jun 23 '25 at 10:16

1 Answers1

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The answer is so tautological and easy that it seems baffling!
If you call $u$ the fraction $u=\frac{p(x)}{q(x)}$ a polynomial $f(T)\in k(u)[T]$ of lowest degree killing $x$ is $$f(T)=p(T)-uq(T)$$ If you want a monic polynomial you will have to divide that polynomial $f(T)$ by some "constant" i.e. by some element in $k(u)$ in order to obtain $\operatorname {min}(x,k(u),T)$.

An example
If $u=\frac {x}{x^2+1} $ you will first compute $f(T)= T-\frac {x}{x^2+1}(T^2+1)=-\frac {x}{x^2+1}T^2+T-\frac {x}{x^2+1}$ and the minimal polynomial of $x$ over $k(u)$ is obtained by dividing by $-\frac {x}{x^2+1}$, so that the required minimal polynomial for $x$ over $k(u)$ is: $$\operatorname {min}(x,k(u),T)=T^2-\frac {x^2+1}{x}T+1=T^2-\frac 1u T+1\in k(u)[T]$$

Georges Elencwajg
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  • why is $f$ irreducible ? (also I think you meant "killing $x$" and not "killing $u$") – mercio Nov 01 '15 at 12:42
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    Yes, I meant killing $x$, thank you. Now corrected. The polynomial is irreducible because it has the right degree ( and the OP knows that degree already). – Georges Elencwajg Nov 01 '15 at 12:53
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    Yes, $[k(x):k(\dfrac{p}{q})]=[k(x):k(\dfrac{p}{q},\dfrac{1}{q})][k(\dfrac{p}{q},1/q):k(\dfrac{p}{q})]\geq\max(\deg(r),\deg(q))$ so the polynomial that Georges introduced has the least possible degree. And so coprimity of $p$ and $q$ are used for irreducibility of this polynomial. – H.W. Nov 01 '15 at 19:14