[Math] About localization of a finitely generated $k$-algebra

algebraic-geometrycommutative-algebra

I've some questions about localization.

My problem is the following.
Let $A$ be a finitely generated domain over a field $k$. We know by hypothesis that its fraction field $K(A)$ is a finite separable extension of a purely trascendental extension $k(T_1, \dots , T_d)$ of $k$ (follows from Noether's Normalization Lemma, right?). Then we can write $K(A)=k(T_1, \dots , T_d)[X]/(P(X))$ where $P$ is irreducible and separable polynomial and without loss of generality we can choose $P(X)$ in $k[T_1, \dots , T_d][X]$ by Gauss' lemma.

Now my book says:

  1. There exists a localization $A_f$ of $A$ such that $k[T_1, \dots , T_d][X]/(P(X)) \subseteq A_f$.
  2. As $A$ is a $k$-finitely generated algebra, there exists $R \in k[T_1, \dots , T_d]$ such that $A \subseteq k[T_1, \dots , T_d]_{R}[X]/(P(X))$.

I don't know how to prove the two assertions above.
Does anyone have some ideas?

Thanks a lot!

PS: If someone is interested in algebraic geometry, this is used by Liu's book to prove that a geometrically reduced variety has a closed regular point.

Best Answer

You only need to use the fact that the rings you're are working with are finitely generated. In the first example, note that by the definition of $P$, $k[T_1,\ldots,T_d][X]/(P(X))$ is just a finitely generated subring of $K(A)$. The second case is almost the same. Consider the generators of $k$-algebra $A$, pick $R$ for each one of them and finally take a product of all $R$'s.

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