Lemma 10.41.11. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume that

1. there exists a unique prime $\mathfrak q \subset S$ lying over $\mathfrak p$, and

2. either

1. going up holds for $R \to S$, or

2. going down holds for $R \to S$ and there is at most one prime of $S$ above every prime of $R$.

Then $S_{\mathfrak p} = S_{\mathfrak q}$.

Proof. Consider any prime $\mathfrak q' \subset S$ which corresponds to a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak p})$. This means that $\mathfrak p' = R \cap \mathfrak q'$ is contained in $\mathfrak p$. Here is a picture

$\xymatrix{ \mathfrak q' \ar@{-}[d] \ar@{-}[r] & ? \ar@{-}[r] \ar@{-}[d] & S \ar@{-}[d] \\ \mathfrak p' \ar@{-}[r] & \mathfrak p \ar@{-}[r] & R }$

Assume (1) and (2)(a). By going up there exists a prime $\mathfrak q'' \subset S$ with $\mathfrak q' \subset \mathfrak q''$ and $\mathfrak q''$ lying over $\mathfrak p$. By the uniqueness of $\mathfrak q$ we conclude that $\mathfrak q'' = \mathfrak q$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

Assume (1) and (2)(b). By going down there exists a prime $\mathfrak q'' \subset \mathfrak q$ lying over $\mathfrak p'$. By the uniqueness of primes lying over $\mathfrak p'$ we see that $\mathfrak q' = \mathfrak q''$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

In both cases we conclude that the map $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) \to \mathop{\mathrm{Spec}}(S_{\mathfrak p})$ is bijective. Clearly this means all the elements of $S - \mathfrak q$ are all invertible in $S_{\mathfrak p}$, in other words $S_{\mathfrak p} = S_{\mathfrak q}$. $\square$

## Comments (6)

Comment #6720 by Zeyn Sahilliogullari on

I think it needs to be explicitly stated that R is a subset of S or that the ring map is an injection/inclusion.

Comment #6726 by Zeyn Sahilliogullari on

I think it needs to be explicitly stated that R is a subset of S or that the ring map is an injection/inclusion. For the trivial reason that the multiplictive set S - 𝔭 makes no sense since 𝔭 is contained in R and not S.

Comment #6727 by on

@#6726. The nonsensical expression $S \setminus \mathfrak p$ is not used anywhere, I think. The notation $S_\mathfrak p$ indicates the localization of $S$ at $\mathfrak p$ viewed as an $R$-module. This localization is an $S$-algebra as it is also the localization of $S$ at a multiplicative subset of $S$, namely image of the set $R \setminus \mathfrak p$ in $S$. (This notation first occurs in Remark 10.18.5.) OK?

Comment #6757 by Zeyn Sahilliogullari on

Yes, the statement was clear after you explained the notation. I could not follow the last two sentences in the proof so I worked it out a different way (this part turns out to be exercise 3.8V in A&M's commutive algebra book).

Comment #6916 by on

OK, I am going to leave it as is. If there is a suggestion for a clearer formulation of the last two sentences, then I am happy to hear it.

There are also:

• 7 comment(s) on Section 10.41: Going up and going down

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