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So basically I have the following: Let $F$ be a field, and let $f(t),g(t)\in F[t]$. Let $u=\frac{f}{g}$ be such that $u\in F(t)\backslash F$, and furthermore assume that $f$ and $g$ are relatively prime. Prove that $[F(t):F(u)]$ is finite and it is equal to $\max\{\deg(f),\deg(g)\}$.

The first part is easy since we can define $h(x)\in F(u)[x]$ by $h(x)=u\cdot g(x)-f(x)$. This is a polynomial with coefficients in $F(u)$ since $g$ and $f$ have coefficients in $F$. Note that $h(t)=0$, hence we see that $t$ is algebraic:$$\Rightarrow [F(t):F(u)]<\infty$$Now in order to calculate the degree I have to either find the minimal polynomial of $t$ over $F(u)$, or find a $F(u)$-basis for $F(t)$ and count its dimension. I believe that $h$ might be the minimal polynomial but I cannot see why it is irreducible. Also, I tried looking for a basis, but I could not come up with one.

This is not homework, so you can go ahead and explain as thoroughly as you want.


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Your "first part" should have $h(x) = ug(x) - f(x)$, not $p(x)$. – Arturo Magidin Apr 13 '12 at 21:02
This is sort of a duplicate of… – Bruno Stonek Nov 20 '12 at 7:05
up vote 4 down vote accepted

To show that $h(x)$ is in fact a (constant multiple of the) minimal polynomial of $u$, use Gauss's Lemma: since $\gcd(g(x),f(x))=1$, the polynomial is primitive when considered as an element of $(F[u])[x]$. Therefore, it is irreducible in $F(u)[x]$ if and only if it is irreducible in $F[u][x] = F[u,x] = F[x][u]$ (this is a consequence of Gauss's Lemma). But irreducibility in $F[x][u]$ is immediate, since the polynomial is of degree $1$ in $u$.

Thus, $h(x)$ is irreducible, and so has the same degree as the monic irreducible of $t$. It is now easy to verify that $\deg(h)=\max\{\deg(f),\deg(g)\}$, as desired.

(Technically, $h$ may not be the minimal polynomial, because it may not be monic, but you don't need it to be the minimal polynomial, you just need it it to be irreducible)

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Thank you so much! – Daniel Montealegre Apr 13 '12 at 22:22

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