$K \otimes_R A$ is Noetherian ring where $R$ is integral domain with field of fractions $K$ and $A$ an $R$-algebra which is Noetherian ring

Question

Let $R$ be an integral domain , $K$ be the field of fractions of $R$ . Let $A$ be an $R$-algebra such that $A$ is a Noetherian ring . Then $K \otimes_R A $ is also an $R$-algebra , my question is : How to show that $K \otimes_R A $ is a Noetherian ring ? The only thing I can relate to is that $K$ is a flat $R$-module , but I don't know whether this has any relevance or not . Please help . Thanks in advance

Answer 1

Let $I$ be an ideal of $K\otimes_RA$, denote by $i:A\rightarrow K\otimes_RA$ defined by $i(a)=1\otimes a$, $i^{-1}(I)$ is an ideal of $A$.

Let $x={p_1\over q_1}\otimes a_1+...+{p_n\over q_n}\otimes a_n, p_i,q_i\in R, a_i\in I$ an element of $I$. Write $q=q_1...q_n$, $qx\in i(i^{-1}(I))$. This implies that $S^{-1} i(i^{-1}(I))=I$ where $S=R-\{0\}$. Consider a family of ideals $I_1\subset I_2...\subset I_n...$ since $A$ is Noetherian, there exists $N$ such that $i^{-1}(I_n)=i^{-1}(I_N), n>N$. $i^{-1}(I_n)=i^{-1}(I_N), n>N$. We deduce that $I_n=S^{-1}i(i^{-1}(I_n))=I_NS^{-1}i(i^{-1}(I_N)), n>N$.