Showing that all elements of a Field Extension are transcendental

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.