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As part of lemma 6.4 in Hartshorne, I came across a statement that I can't prove

Let $m,n $ be maximal ideals of an integral domain $A$. Then $ A_m \subset A_n$ implies $n \subset m $.

It is clear intuitively. To prove it, I picked $s \in A-m$. Since $1/s\in A_m \subset A_n$, there is $a/t=1/s$ for $a\in A,t\in A-n$. This gives $as=t$. I don't know how to use this to show that $s\in A-n$.

I also wonder whether we can generalise this to any multiplicative sets or any rings (not domains).

Thank you

user26857
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PeterM
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1 Answers1

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$as=t \notin n$ implies $s \notin n$ since $n$ is an ideal.

In general, if $A$ is a commutative ring and $S,T \subseteq A$ are multiplicative subsets, then there is a homomorphism of $A$-algebras $A_S \to A_T$ (and this is unique) if and only if $A \to A_T$ sends the elements of $S$ to units, if and only if for all $s \in S$ there is some $a \in A$ and $t \in T$ such that $\frac{s}{1} \cdot \frac{a}{t}=1$ in $A_T$, i.e. $\frac{sa}{t}=1$, i.e. there is some $t' \in T$ such that $t'(t-sa)=0$. Hence, $A_S \to A_T$ exists iff for all $s \in S$ there is some $a \in A$ such that $sa \in T$.

In the special case $S=A \setminus \mathfrak{p}$, $T = A \setminus \mathfrak{q}$ for two prime ideals $\mathfrak{p},\mathfrak{q}$, we see that there is a homomorphism of $A$-algebras $A_\mathfrak{p} \to A_\mathfrak{q}$ iff $A \setminus \mathfrak{p} \subseteq A \setminus \mathfrak{q}$ iff $\mathfrak{q} \subseteq \mathfrak{p}$.

Martin Brandenburg
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