# For a finitely generated $K$ algebra $A$ is $S^{-1}A$ also finitely generated $K$ algebra?

Let $A$ be a finitely generated $K$ algebra and $S \subset A$ be any multiplicative closed set. Then can I say $S^{-1}A$ is also finitely generated $K$ algebra ? In particular my interest is when $A$ is Artinian ring and $S$ is complement of a maximal ideal.

My Try: If $A=K[X_1,\ldots,X_n]/I$ and $S \cap I=\phi$ then $S^{-1}A \cong S^{-1}K[X_1, \ldots,X_n]/ S^{-1}I$ after that I cannot conclude !

• Try $\Bbb C[x]$ and it's field of fractions $\Bbb C(x)$ – xsnl Mar 28 '18 at 9:44
• I guess since there are infinitely many non associate primes in $\mathbb{C}[X],$ $\mathbb{C}(X)$ is not finitely generated $\mathbb{C}[X]$ algebra. – user371231 Mar 28 '18 at 9:51
• Is it true if $A$ is Artinian ring and $S$ is complement of a maximal ideal ? – user371231 Mar 28 '18 at 9:52

Yes, if $A$ is a finitely generated Artinian algebra over $K$ and $S=A\setminus \frak m$ is the complement of a maximal ideal, then $S^{-1}A$ is finitely generated over $K$. Here is why:
If the maximal ideals of $A$ are $\mathfrak m_1=\mathfrak m,\cdots,\mathfrak m_r$ we have $A\cong A/\mathfrak m_1^N\times \cdots A/\mathfrak m_r^N$, where $N$ is a sufficiently large integer [Atiyah-Macdonald, Theorem 8.7], so that $$S^{-1}A\cong S^{-1}(A/\mathfrak m_1^N)\times \cdots S^{-1}(A/\mathfrak m_r^N)\quad (\bigstar)$$ a) Now, for $i\geq 2, S^{-1}(A/\mathfrak m_i^N)=S^{-1}A/S^{-1}\mathfrak m_i^N=0$ since $S^{-1}\mathfrak m_i^N=S^{-1}A$ because $\mathfrak m_i^N$ meets $S$ [ Atiyah-Macdonald, Proposition 3.11 ii)].
b) On the other hand $S^{-1}(A/\mathfrak m_1^N)=A/\mathfrak m_1^N$ because every $s\in S$ has an image $\bar s\in A/\mathfrak m_1^N$ which is already invertible as a consequence of not belonging to $\mathfrak m_1/\mathfrak m_1^N$, the only maximal ideal of the local ring $A/\mathfrak m_1^N$.
In view of a) and b), $(\bigstar)$ implies $S^{-1}A\cong A/\mathfrak m_1^N$, thus proving that $S^{-1}A$ is finitely generated over $K$, because it is a quotient of $A$.