# Showing that all elements of a Field Extension are transcendental

My question is whether my strategy for this proof is correct. I'll put the fact to be proven, my strategy, and why I am hesitant about my strategy.

Thing to be proven: I want to show that if $$E$$ is an extension field of $$F$$, and $$\alpha \in E$$ is transcendental over $$F$$ then every element of $$F(\alpha)$$ (field of quotients of $$F[\alpha]$$) is also transcendental over $$F$$.

My Proof Strategy: Assume $$\beta \in F(\alpha)$$ and $$\beta$$ is algebraic. Then $$\exists p(x) \in F[x]$$ such that $$p(\beta) = 0$$. But since $$\beta$$ is a polynomial in $$\alpha$$, that means that some polynomial in $$\alpha$$ with coefficients in $$F$$ is equal to $$0$$ so $$\exists q(x) \in F[x]$$ such that $$q(\alpha) = 0$$.

My Concerns:

• $$\beta$$ is in a field of quotients of $$F[\alpha]$$, but this can be circumvented because in order for that to be $$0$$ we just need the "numerator" which is in $$F[\alpha]$$ to have a root.

• Showing rigorously that $$p(\beta)$$ is a polynomial in $$\alpha$$ seems like a pain. It makes sense, but doing a general solution seems hard.

This makes me think that an easier solution might be to somehow employ theorems about the fact that $$\langle p(x)\rangle$$ must be a maximal ideal.

So my question is, which route would yield an easier and more elegant solution?

• What is $E$ and how does it enter in to this question? Cheers! – Robert Lewis Nov 2 '18 at 2:52
• @RobertLewis thanks for pointing that out, it should be $\alpha \in E$ – SalmonKiller Nov 2 '18 at 2:57

Proffering the use of uniqueness of factorization of polynomials as an alternative.

Assume that contrariwise that $$\beta=r(\alpha)/s(\alpha), \beta\notin F,$$ is algebraic over $$F$$. Here $$r(\alpha),s(\alpha)\in F[\alpha]$$. W.l.o.g. we can assume that:

• $$\gcd(r(\alpha),s(\alpha))=1$$ because we can simply cancel any common factor, and
• at least one of $$r(\alpha), s(\alpha)$$ is not a constant. For otherwise $$\beta\in F$$.

Let $$p(x)=a_0+a_1x+\cdots+a_nx^n\in F[x]$$ be the minimal polynomial of $$\beta$$ over $$F$$. Because $$\beta\notin F$$ we know that $$p(x)$$ has degree at least two. Furthermore, $$p(x)$$ is irreducible in $$F[x]$$, so we can assume that $$a_n=1$$ and $$a_0\neq0$$.

Let's plug in $$x=\beta$$. We arrive at $$a_0+a_1\frac{r(\alpha)}{s(\alpha)}+\cdots+\left(\frac{r(\alpha)}{s(\alpha)}\right)^n=0.$$ Using a common denominator allows a rewrite $$0=\frac{a_0s(\alpha)^n+a_1s(\alpha)^{n-1}r(\alpha)+a_2s(\alpha)^{n-2}r(\alpha)^2+ \cdots+a_{n-1}s(\alpha)r(\alpha)^{n-1}+r(\alpha)^n}{s(\alpha)^n}.$$ Here the numerator must be the zero polynomial in the ring $$F[\alpha]$$. But, the two bullets above imply that there exists an irreducible non-constant polynomial $$m(\alpha)\in F[\alpha]$$ such that $$m(\alpha)$$ is a factor of exactly one of $$r(\alpha), s(\alpha)$$ (and hence not a factor of the other).

Look at the numerator again. If $$m(\alpha)\mid r(\alpha)$$ and $$m(\alpha)\nmid s(\alpha)$$, then all the terms in the numerator with the sole exception of the first are divisible by $$m(\alpha)$$. Implying that the numerator is not divisible by $$m(\alpha)$$. This is a contradiction because the numerator is supposed to be zero.

If $$m(\alpha)\mid s(\alpha)$$ and $$m(\alpha)\nmid r(\alpha)$$ then a similar problem occurs. This time the last term of the numerator is the only one not divisible by $$m(\alpha)$$.

Unique factorization of polynomials was used in the step when we deduced that a power of a polynomial not divisible by $$m(\alpha)$$ could not itself be divisible by $$m(\alpha)$$ either.

• Not sure whether this should be called "elegant" :-) – Jyrki Lahtonen Nov 1 '18 at 10:22
• Are you sure that we need to use the uniqueness of factorization in this case? Since $m(\alpha)$ is irreducible and $F[\alpha]$ is an integral domain, doesn't that imply that the result that you are using unique factorization for is automatically provided by properties of Integral Domains? – SalmonKiller Nov 1 '18 at 17:44
• @SalmonKiller Possibly :-) All I need is that $F[\alpha]/\langle m(\alpha)\rangle$ has no nilpotent elements. – Jyrki Lahtonen Nov 1 '18 at 18:29

I would use the multiplicativity of the degree in the tower of extensions $$F \subseteq F(\beta) \subseteq F(\alpha)$$. Since $$\beta \in F(\alpha)$$, then $$\beta$$ is a rational function in $$\alpha$$, i.e., $$\beta = \frac{f(\alpha)}{g(\alpha)}$$ for some polynomials $$f,g \in F[x]$$, $$g \neq 0$$. Then $$\alpha$$ is a root of $$\beta g(x) - f(x) \in F(\beta)[x]$$. Assume $$\beta \in F(\alpha) \setminus F$$. Then at least one of $$f$$ and $$g$$ is nonconstant, and since $$F[x]$$ is a PID we may take $$f$$ and $$g$$ relatively prime. Then $$\beta g(x) - f(x)$$ is not the zero polynomial (since $$\beta g \in F(\beta)[x] \setminus F[x]$$ while $$f \in F[x]$$), so $$\alpha$$ is algebraic over $$F(\beta)$$, hence $$[F(\alpha) : F(\beta)] < \infty$$.

For contradiction, assume $$\beta$$ is algebraic over $$F$$. Then $$[F(\beta) : F] < \infty$$, so by multiplicativity we have $$[F(\alpha) : F] = [F(\alpha) : F(\beta)] [F(\beta) : F] < \infty \, ,$$ contradicting the fact that $$\alpha$$ is transcendental over $$F$$.

• hey André, thanks for answering so quickly. I was wondering if you could chew it out a bit more, since it seems to me like concepts of the tower of extensions hide some some more fundamental underlying patterns. And since the book that I'm following along talks about the "tower of extensions" in full detail some time later, it seems like this would be possible with more fundamental concepts that don't hide the internals with abstraction. Not saying your proof is wrong, but simply asking for more "clarity". Thanks! – SalmonKiller Nov 1 '18 at 4:49
• The basic fact that I'm using is that if $K/F$ is a field extension, then $\theta \in K$ is algebraic over $F$ iff $[F(\theta) : F]$ is finite. This avoids the problem in your second bullet point, since we can check that a number is finite, rather than trying to manipulate polynomials. I doubt there is an elegant way working directly with polynomials, but I could be wrong. – André 3000 Nov 1 '18 at 5:04
• In any case, you shouldn't be turned away too much by the multiplicativity of degree in a tower: it basically comes down to counting elements in a basis, as shown here and here. – André 3000 Nov 1 '18 at 5:04
• That Why is it clear that $\beta g-f$ is not the zero polyomial? That would imply $\beta g(0)-f(0)=0$ and as we can assume wlog that one of $f,g$ is not a multiple of $x$, we'd arrive at $\beta\in F$ – Hagen von Eitzen Nov 1 '18 at 5:51
• @HagenvonEitzen I see: I think I need to assume $\beta \notin F$ at that point. Since $F[x]$ is a UFD, then can choose $f$ and $g$ relatively prime and not both constant. I'll edit. – André 3000 Nov 1 '18 at 5:59