0

I am aware that questions related to the title of this post have already been posted but I think that the proof I'm looking for is a bit different. This proof uses the two following results which I was not able to prove either:

1) If $L$ is a finite algebraic extension of $\mathbb{Q}$ and $R \subseteq L$ a finitely generated $\mathbb{Z}$-algebra, there exist positive integers $n_1,...,n_k$ such that $R$ is integral over $\mathbb{Z}[\frac{1}{n_1},...,\frac{1}{n_k}]$.

2) If $\mathfrak{m}$ is a maximal ideal of $\mathbb{Z}[X_{1},...,X_k]$ it must contain a prime number $p\in \mathbb{Z}$.

And then deduce the theorem from the title.

user26857
  • 52,094
WrabbitW
  • 547
  • $R=\mathbb Z[x_1,\dots,x_k]$; since $x_i\in L$ it's algebraic over $\mathbb Q$, so there is $n_i>0$ such that $n_ix_i$ is integral over $\mathbb Z$.
    • – user26857 May 18 '17 at 06:52
  • $m\cap\mathbb Z=(0)\implies\mathbb Q\subset\mathbb Z[X_1,\dots,X_k]$ is finitely generated. Now use the Zariski's lemma. (This is certainly on this site.)
    • – user26857 May 18 '17 at 06:56
  • 1
  • I don't think your approach is different from this.
    • – user26857 May 18 '17 at 06:59