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Problem: Suppose that $K$ is a field and that $f$ and $g$ are relatively prime in $K[x]$. Show that $f- yg$ is irreducible in $K(y)[x]$.

My attempt: Consider the polynomial $f-gY\in (K[x])[Y]$. This polynomial is irreducible. Indeed, if it were not, since its degree is over $(K[x])[Y]$ is $1$, $f-yg=(a-by)c$ where $a, b, c\in K[x]$ and the degree of $c$ over $K[x]$ is at least $1$. Then $f-ac=(g-bc)Y$, therefore $f=ac$ and $g=bc$ which contradicts $f$ and $g$ relatively prime.

Now, because $f-gY$ irreducible, if $y$ transcendental over $K$, $f-gy$ will be transcendental over $V[x]$, where $V$ is the k-vectorial space generated by $\lbrace y^n; n\in \mathbb{Z} \rbrace \cup \lbrace 1\rbrace$.

Finally, by Gauss's lema $f-gy$ is irreducible over $K(y)[x]$, since $K(y)$ is the fraction field of $V$.

I don't know if my proof is correct or not... moreover, I don't know how I can prove the statament when $y$ is algebraic.

Comments: This exercise is 5.4 of the book a course in Galois theory by Garling.

3435
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    This is obviously false if $y$ is algebraic. Take $f=x+1,g=x,y=1$. – GreginGre May 07 '23 at 10:27
  • @GreginGre but in this case $f-yg=1$ and $1$ is an irreducible element... – 3435 May 08 '23 at 02:03
  • No, it isn't. An irreducible element is by definition non invertible. (or, if you prefer to stick to the context of your question: an irreducible polynomial of $K[X]$ is non constant). – GreginGre May 08 '23 at 07:50

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