Does $P\subset Q$ imply $A_P \subset A_Q$?

Question

P and Q are prime ideals of A such that $P\subset Q$
Whether $A_P \subset A_Q$ or $A_Q\subset A_P$? ($A_P$ and $A_Q$ are localization at P and Q respectively )
I think it will be $A_Q\subset A_P$. As $P\subset Q$ will imply $P^c \supset Q^c$, so $A_P$ will have more elements than $A_Q$. But I am not quite sure. Any help?

Answer 1

You can use the uninversal property of localization to get your answer here. (In the following, all rings are commutative with identity, and ring homomorphisms preserve the identity.)

Universal property: Let $A$ be a ring, and let $S\subseteq A$ be a multiplicative subset. Then there exists a ring $S^{-1}A$ and a morphism $\ell : A\to S^{-1} A$ such that given a ring homomorphism $f : A\to B$ such that $f$ sends $S$ to units in $B$ (i.e., $f(S)\subseteq B^\times$), there exists a unique ring homomorphism $f' : S^{-1}A \to B$ such that $f$ factors as $f = f'\circ \ell.$

Observe that this implies that every element of $S$ is mapped to a unit in $S^{-1}A$ by $\ell.$

For a prime ideal $\mathfrak{P}\subseteq A,$ $S_\mathfrak{P}:=A\setminus\mathfrak{P}$ is a multiplicative subset, and $A_\mathfrak{P} := S_\mathfrak{P}^{-1} A.$ It follows that if $\mathfrak{P}\subseteq\mathfrak{Q},$ then $S_\mathfrak{Q}\subseteq S_\mathfrak{P}$. The canonical maps $\ell_\mathfrak{P} : A\to A_\mathfrak{P}$ and $\ell_\mathfrak{Q} : A\to A_\mathfrak{Q}$ map $S_\mathfrak{P}$ and $S_\mathfrak{Q}$ to units, so $S_\mathfrak{Q}\subseteq\mathfrak{P}$ implies that $\ell_\mathfrak{P}(S_\mathfrak{Q})\subseteq A_\mathfrak{P}^\times.$ Thus, the universal property of localizations implies that there exists a unique map $f$ such that the canonical map $\ell_\mathfrak{P}$ factors as $$ A\xrightarrow{\ell_\mathfrak{Q}} A_\mathfrak{Q}\xrightarrow{f} A_\mathfrak{P}. $$

Essentially, the above boils down to what you said in your question: $S_\mathfrak{P}$ contains $S_\mathfrak{Q}$, so when you localize at $\mathfrak{P}$ you invert more elements than when localizing at $\mathfrak{Q}.$ Since you need to invert every element of $S_\mathfrak{Q}$ in order to invert every element of $S_\mathfrak{P},$ you can get to $A_\mathfrak{Q}$ to $A_\mathfrak{P}$ by a further localization (i.e., inverting more elements).

However, one should remark that it is not necessarily true that the map $A_\mathfrak{Q}\to A_\mathfrak{P}$ will be an inclusion. For example, let $k$ be a field, and let $A = k[\![x,y]\!]/(x^2,xy).$ This is a local ring with maximal ideal $\mathfrak{m} = (x,y)A,$ and we can see that $\mathfrak{p} = xA\subseteq \mathfrak{m}$ is also prime: $$ A/\mathfrak{p}\cong k[\![x,y]\!]/(x^2,xy,x)\cong k[\![x,y]\!]/(x)\cong k[\![y]\!]. $$ If we localize $A$ with respect to $\mathfrak{m},$ the canonical map $A\to A_\mathfrak{m}$ is an isomorphism. However, the map $A\to A_\mathfrak{p}$ is not injective: we have $$ \begin{align*} A_\mathfrak{p}&\cong k[\![x,y]\!]_{\mathfrak{p}}/(x^2,xy)_\mathfrak{p}\\ &\cong k[\![x,y]\!]_{\mathfrak{p}}/(x^2,x)_\mathfrak{p}&\textrm{(because }y\not\in\mathfrak{p},\textrm{so $y$ becomes a unit in }k[\![x,y]\!]_{\mathfrak{p}}\textrm{)}\\ &\cong k[\![x,y]\!]_{\mathfrak{p}}/(x)_\mathfrak{p}. \end{align*} $$ But now, we have $$ k[\![x,y]\!]_{\mathfrak{p}}/(x)_\mathfrak{p} \cong \left(k[\![x,y]\!]/(x)\right)_\mathfrak{p}\cong k[\![y]\!]_\mathfrak{q}, $$ where $\mathfrak{q}$ is the zero ideal in $k[\![y]\!]$ ($xA$ becomes the zero ideal of $k[\![y]\!]).$ Thus $$ A_\mathfrak{p}\cong\operatorname{Frac}k[\![y]\!], $$ and the map $A\cong A_\mathfrak{m}\to A_\mathfrak{p}\cong\operatorname{Frac}k[\![y]\!]$ maps $x$ to $0.$

To summarize, given an inclusion of prime ideals $\mathfrak{P}\subseteq\mathfrak{Q}$ of some ring $A,$ we obtain a canonical ring homomorphism $A_\mathfrak{Q}\to A_\mathfrak{P},$ but writing $A_\mathfrak{Q}\subseteq A_\mathfrak{P}$ is somewhat misleading, as the canonical ring homomorphism mentioned is not necessarily an injection.