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While trying assignment questions of Field Theory of my class I am unable to solve this particular problem.

Let $ f/g \in K(x) $ with f/g not belonging to K and f, g a relatively prime in K[x] and consider the extension of K by K(x).

Then prove that x is algebraic over K(f/g) and [ K(x) : K(f/g) ] = max ( deg f, deg g).

I have showed it algebraic by considering polynomial p(a)= f/g × g(a) - f(a).

But I am unable to prove the result related to degree.

Edit 1: This question had some part which I am unable to solve : The assignment $ x \to f/g $ induces a homomorphism $\sigma : K(x) \to K(x)$ such that $\frac{\phi(x)}{\psi(x)} \to \frac{\phi(f/g)}{\psi(f/g)}$.

I am not well in problems in which it is to be proved that an homomorphism is induced. Can you please tell basic philosophy behind it?

Edit : So , my question has been linked to the question here:Let $x$ be transcendental over $F$. Let $y=f(x)/g(x)$ be a rational function. Prove $[F(x):F(y)]=\max(\deg f,\deg g)$

But as I was reading following question I have 2 questions in 1st two lines of attempt of user: (i) Why does OP can and want to replace y by 1/y (ii) In 1st line itself OP writes that deg g≥deg f. But then how by the Euclid algorithm OP assumed we assume deg g>deg f?

These are some of my question by which I am not able to understand the answer and attempt and I request you that can you please clear these questions of mine or answer it separately in case you think that is required.

Thanks!!

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    For the homomorphism induced you only need to show that the definition makes sense; i.e. $f/g$ is transcendental over $K$ (so $\psi(f/g)\neq 0$). It's the same idea as the proof that all transcendental extensions are isomorphic (to $K(x)$). For you other problem see here. Hope this helps! – August Liu Nov 03 '20 at 10:29
  • This is a pretty standard result and a precursor to Lüroth's theorem discussed here: https://math.stackexchange.com/q/4237055/72031 The irreducibilty of the polynomial $yg(x) - f(x) $ is first established over $K[x, y] $ and then over $K(y) [x] $ where $y=f(x) /g(x) \in K(x) \setminus K$. The proof uses Gauss Lemma. – Paramanand Singh Sep 17 '21 at 09:55

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