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Suppose that $F$ is a field and that $u \in F(x):= \{PQ^{-1}:P,Q \in F[x], Q\neq 0 \}$, so that $F \subseteq F(u) \subseteq F(x)$. Is there a general method for determining $[F(x):F(u)]$?

For my homework problem I have been given the specific case of $u := \frac{x^3}{x+1} = x^2-x+1-\frac{1}{x+1}$ and we haven't even gone over anything remotely similar to a problem like this in class. The hint was to choose $v \in F(x)$ so that $F(u,v) = F(x)$.

If we had such a $v$ then $[F(x):F(u)] = [(F(u))(v):F(u)]$ which is equal to the minimum degree of a polynomial with coefficients in $F(u)$ which has $v$ as a root. But I don't know how to find such a polynomial, nor is it obvious that one exists.

If we take $v:=x$ then certainly $F(u,x)=F(x)$ but I didn't make any progress from this. I was thinking of writing a Mathematica script to try random polynomials but I don't think this is the intended method of solution.

I was also thinking that it might be significant that the roots of the numerator and denominator of $u$ are 0 and 1, namely they are elements of $F$.

Any help would be appreciated.

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2 Answers 2

up vote 4 down vote accepted

Hans gives you a method to proving $[F(x):F(u)]=3$ in your particular case. Here is the general answer. Write $u=P(x)/Q(x)\in F(x)$ with $\gcd(P, Q)=1$. Then $$ [F(x): F(u)]=\max\{ \deg P(x), \deg Q(x) \}.$$

Proof. As $F(u)=F(1/u)$, one can suppos $\deg P\ge \deg Q$. By changing $u$ with $u-c$ for a suitable $c\in F$, we can suppose $d=\deg P>\deg Q$. Consider the polynomial $$ H(T)=P(T)-uQ(T)\in F(u)[T].$$ It has degree $d$, and vanishes at $x\in F(x)$. It remains to show that $H(T)$ is irreducible in $F(u)[T]$.

Note that $H(T)\in F[u][T]$. By Gauss Lemma, it is enough to show that $H$ is an irreducible element of the UFD ring $F[u][T]$. But $F[u][T]=F[u,T]=F[T][u]$. Viewed as a polynomial in coefficients in $F[T]$, $H$ has degree $1$, and its coefficients $P(T), Q(T)$ are coprime. So $H$ is an irreducible lement of $F[u,T]$ and we are done.

Recall that if $A$ is an UFD, the irreducible elements of $A[T]$ are exactly the irreducible elements of $A$, and the polynomials $R(T)\in A[T]$ (of positive degree) whose coefficients are coprime and such that $R(T)$ is irreducible in $\mathrm{Frac}(A)[T]$.

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This is exactly what I was looking for. One question though. Why do we need that $P(T),Q(T)$ are coprime? Is it not true that any polynomial of degree 1 is irreducible? I see that it must be necessary because if $P,Q$ are not coprime then we would get a different answer... But I don't see where it is used. –  nullUser Nov 16 '11 at 21:59
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Without the hypothesis on $\gcd(P(T), Q(T))$, then $H$ is only irreducible in $F(T)[u]$, not in $F[T][u]$ because it is divisible by the gcd which is not invertible in $F[T][u]$. –  user18119 Nov 16 '11 at 22:47
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Observe $x$ satisfies the polynomial $f(y) = y^3 - \left(\frac{x^3}{x + 1}\right)(y + 1) \in F(u)[y]$. This demonstrates $[F(x):F(u)] \leq 3$. You should be able to show $[F(x):F(u)] \neq 1$ by arguing $x \not \in F(u)$. I'll leave eliminating $[F(x) : F(u)] \neq 2$ to you.

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How would one argue that $x \notin F(u)$? Suppose that $$x=\frac{n_0 + n_1u + \ldots + n_ku^k}{d_0 + d_1u + \ldots + d_rk^r}.$$ I don't see how to derive a contradiction from this, and I would imagine that showing $[F(x):F(u)]\neq 2$ is even more complicated. Is there a simpler way that what I am proposing? –  nullUser Nov 16 '11 at 20:08
    
@Kb100: Actually, I've not quite figured that out. I've upvoted QiL's answer, as it's more along the lines of what you're looking for anyway. It's also clearly more thought out. –  Hans Parshall Nov 16 '11 at 23:19
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