field extension and equivalent automorphisms

Question

Suppose we have a field extension $L\supseteq K$ and two automorphisms $\sigma,\tau\in Gal(L:K)$ that are equal for every generator $\alpha_{i}$ (i.e $\sigma(\alpha_{i})=\tau(\alpha_{i}),\forall i)$ then $\sigma(\alpha)=\tau(\alpha)$ for every $\alpha\in L$

I don't understand the following step in the proof:

Let $L=K(\alpha_{1},......,\alpha_{r})$ if $\alpha\in L$ then $\exists$ polynomials $f,g\in K[x_{1},x_{2},.....,x_{r}]$ st

$\alpha=\frac{f(\alpha_{1},...\alpha_{r})}{g(\alpha_{1},..,\alpha_{r})}$ with non-zero denominator

How do we get this fraction? How do we know that? and what is the polynomial ring above? Ring in many variables?

I take it as all $\alpha_{1},\alpha_{2},...,\alpha_{4}$ are roots for some some polynomials? what is then $\alpha$?

Thanks

Answer 1

If $\alpha\in L$, then it can be written as algebraic combinations of your $\alpha_i$. That is where you get your fraction from. Yes, your polynomials are multivariate polynomials in $r$ variables. Your $\alpha_i$ are algebraic over your bottom field and hence are roots of their respective minimal polynomial. The $\alpha$ is just any element of $L$. To show two functions are equal you show that they are equal elemtwise. That is why you are considering such an $\alpha$.

Answer 2

Since $L = K(\alpha_1, \dots, a_r)$ every element of $L$ can, by definition, be written as the quotient of two polynomial expressions over $K$ in $\alpha_1, \dots, a_r$. This is what the step is formally expressing.

The fact that the $\alpha_i$ are roots of some polynomial over $K$ is irrelevant. And $\alpha$ is just an arbitrary element of $L$ of which you're going to show that $\sigma(\alpha) = \tau(\alpha)$.