[Math] Showing that all elements of a Field Extension are transcendental

abstract-algebraextension-fieldfield-theory

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?

Best Answer

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.

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