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Let $F$ be a field and $x$ transcendental over $F$. Let $y=f(x)/g(x)$ be a nonconstant rational function with relatively prime polynomials $f,g\in F[x]$. Prove $[F(x):F(y)]=\max(\deg f,\deg g)$.

My attempt: By replacing $y$ with $\frac{1}{y}$ if necessary we assume $\deg g\geq\deg f$. By the Euclid algorithm we assume $\deg g>\deg f$. Then $\deg g=\max(\deg f,\deg g)$, let $n=\deg g$. Let $R(t)$ be this polynomial $$R(t)=yg(t)-f(t)=\frac{f(x)}{g(x)}g(t)-f(t)$$ Then $\deg R=\deg g=n$ because $n=\deg g>\deg f$. The coefficients of $R(t)$ lie in $F(y)$ and hence $R(t)\in F(y)[t]$. Clearly $R(x)=0$.

Now I only have to prove $R(t)$ is actually the minimal polynomial of $x$ over $F(y)$. I believe the next step is to obtain a contradiction that $R(t)$ being reducible can lead to $f,g$ being not coprime, but I can't figure out how to do that.

P.S. A hint is preferred over a full answer.

trisct
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  • what did you exactly meant by "y replacing y with 1/y "in 1st line of ypur attempt. Can you please explain? –  Sep 17 '21 at 05:32
  • How can it be proved that the Euclid algorithm implies that we assume deg g>deg f? Can you please explain ? –  Sep 17 '21 at 05:48
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    @No-One To your 1st comment: We are considering field extensions, so it doesn't matter whether you extend $F$ with $y$ or $1/y$, i.e. $F(y)=F(1/y)$, since a field must contain the inverse of any nonzero element. To your 2nd comment: Now that we've assumed $\deg g\geq\deg f$, either their degrees are equal or not. If equal, then $f(x)=cg(x)+r(x)$, where $c\in F$ is the quotient of the coefficients of the highest-degree terms in $f(x)$ and $g(x)$, and $\deg r<\deg g$. So $y=c+r(x)/g(x)$. Again, $c\in F$ doesn't matter in field extensions, i.e., $F(y)=F(y-c)$ for $c\in F$. – trisct Sep 17 '21 at 07:36
  • I understood your comment completely that if degrees are equal then f(x)= c g(x) + r(x), then y can be written as equal to c +r(x)/g(x). But you didn't concluded your comment. –  Sep 20 '21 at 11:40
  • what were you trying to prove? –  Sep 20 '21 at 11:48
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    @No-One The conclusion is that "we can assume $\deg f<\deg g$ in the problem statement." – trisct Sep 20 '21 at 12:42
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    @No-One More precisely, we have these equivalences: (i) $[F(x):F(y)]=\max(\deg f,\deg g)\iff[F(x):F(1/y)]=\max(\deg f,\deg g)$. (ii) When $f=cg+r$ as above, $[F(x):F(y)]=\max(\deg f,\deg g)\iff[F(x):F(y-c)]=\max(\deg r,\deg g)$. Note that these equivalences are obtained by plain notation substitution. – trisct Sep 20 '21 at 12:54
  • " Got it. Thanks!! –  Sep 21 '21 at 07:07

1 Answers1

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Hint : Since $x$ is transcendental over $F$, we get that $y$ is also transcendental. In particular, $F[y]$ is a UFD, therefore to show that $R$ is irreducible in $F(y)[t]$, it is enough to show that it is irreducible in $F[y][t]$ by Gauss' lemma.

But then $F[y][t] = F[t][y]$, so you can act like $y$ is a variable and we can take the $y$ degrees during comparison of equality in this integral domain. What is the $y$ degree of $R(t)$?

To finish, what is the $t$ degree of $R$?

Sarvesh Ravichandran Iyer
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  • Got it. If $R(t)$ has a factorization in $F[y,t]$ then one of the factors must have degree $0$ in $y$. Divide $R(t)$ by this factor one yields a polynomial, meaning this factor is common factor of $f,g$. But they are coprime hence the factor must be a constant. Am I right? – trisct Sep 12 '19 at 10:41
  • Yes, that is absolutely fine! – Sarvesh Ravichandran Iyer Sep 12 '19 at 12:21
  • @TeresaLisbon Kindly Consider the question of mine to which this question was linked: https://math.stackexchange.com/questions/3892330/finding-degree-of-field-extension . So, you are giving the proof that R(t) is actually minimal polynomial of x over F(y).So, your answer starts by using that x is transcendental over F. But In my question such is not the assumption. So, Can you please tell how should I proceed? I understood most of attempt of user: trisct. –  Sep 20 '21 at 11:55
  • @No-One Will take a look, thanks. Possibly not today, but tomorrow if possible. – Sarvesh Ravichandran Iyer Sep 20 '21 at 11:56
  • @TeresaLisbon It's Ok. No hurries. –  Sep 20 '21 at 11:56