How to prove this preliminary result for Noether Normalization Lemma?

Question

I need to prove that for an arbitrary field $K$, if $f \in K[x_1,\ldots,x_n]$ is a non-zero polynomial, then there exist $a_1,\ldots,a_{n-1} \in \mathbb N$ and $\lambda \in K$ such that if $$ f(y_1+y_n^{a_1},\ldots,y_{n-1}+y_n^{a_{n-1}},y_n)=g \in K[y_1,\ldots,y_n] $$ then $\lambda g$ is monic as polynomial of $y_n$.

Answer 1

(I came to an answer by rubber-ducking, and since I do not find a similar answer on MathSE I'll leave my result here)

This basically amounts to check that the terms of $g$ involving $y_n$ alone do not all end up with a zero coefficient, or somewhat cancel out. In particular, since all the terms of $f$ are of the form $$ c_{k_1,\ldots,k_n} x_1^{k_1}\cdot\ldots\cdot x_n^{k_n} $$ expanding the binomials which arise from the substitution given, we get terms of the form $$ c_{k_1,\ldots,k_n} y_n^{a_1k_1+\ldots+a_{n-1}k_{n-1}+k_n} $$ plus some other terms which involves more than $y_n$ alone, so that we can ignore them.

Notice we get a term like the above for each of the terms in the original polynomial, so that the leading $y_n$ term of $g$ is the non-null term of highest degree, say $d$. Hence that term is $$ \left( \sum_{a_1k_1+\ldots+a_{n-1}k_{n-1}+k_n=d} c_{k_1,\ldots,k_n} \right)\ y_n^d = \alpha\ y_n^d. $$ Then we want to show we can choose the $a_i$s carefully enough to not have $\alpha$ vanish.

Since we cannot expect to tame all the possible combinations of coefficients, let's pursue a stupid-easy strategy: make the sums $a_1k_1+\ldots+a_{n-1}k_{n-1}+k_n$ all distinct so as to isolate each $c_{k_1,\ldots,k_n}$ to its own term, eliminating any possibility of "cancelation phenomena". Then the first term (i.e. the one with the highest degree) that does not end up with a zero coefficient is our leading term. Observe there's no possibility of having no such term, because we've already noticed each of the coefficients of $f$ will get a $y_n$-only term and since at least one of them is non-zero (because $f\neq 0$), then one of them will be leading.

Therefore we need to find a strategy to choose the $a_i$s in such a way to make all the sums $$ a_1k_1+\ldots+a_{n-1}k_{n-1}+k_n $$ distinct. The key trick here is to see a sum like that as the representation of a number in a positional system in base $N$: the $k_i$s (because they're the variables in our problem) are its 'figures' and the $a_i$s are instead the ascending powers of $N$. To really have a correspondence between the two objects, we need to be sure that the $k_i$s are never bigger than $N$, because in the representation of a number in base $N$ no figures bigger than $N-1$ can be used (otherwise we lose the uniqueness of the representation, which instead is the key property we want to leverage!). But this is an easy task: since the $k_i$s represent the degrees of the indeterminates appearing in the monomials of $f$, which in turn are only in a finite number (or better, appear with a non-null coefficient only finitely many times), we can surely find an $N$ bigger than all of them. Then we've proved our statement, because $\lambda \in K$ can be chosen to be the multplicative inverse of the (non-zero) coefficient of the leading term of $g$, thereby turning it to a $1$.