The form of prime ideals in $S^{-1}R$

Question

Theorem. Let $R$ be a commutative ring with $1_R$, $P\in \mathrm{Spec}(R)$ a prime ideal of $R$, $S\subseteq R$ an multiplicative subset of $R$ and $\nu:R\longrightarrow S^{-1}R,\ a \longmapsto \nu(a):=\frac{a}{1_R}$ the natural homomorphism. Then:

I. a) If $P\in \mathrm{Spec}(R)$ and $P\cap S =\emptyset$, then $P^e\in \mathrm{Spec}(S^{-1}R)$.

b ) If $P'\in \mathrm{Spec}(S^{-1}R)$, then $(P')^c\in \mathrm{Spec}(R)$ and $(P')^c\cap S=\emptyset$.

ΙΙ. Every prime ideal of $S^{-1}R$ is in the form $P^e$, for some unique prime ideal $P\in \mathrm{Spec}(R)$ of $R$ which has the property $P\cap S=\emptyset$. That is, $$\boxed{\mathrm{Spec}(S^{-1}R)=\{ I\in \mathcal{ID}(S^{-1}R): \exists! P\in \mathrm{Spec}(R) \text{ s.t. } I=P^e \text{ and } P\cap S=\emptyset\} } .$$

Proof. I. a) Let $P\in \mathrm{Spec}(R)$ and $P\cap S =\emptyset$. We will show that $P^e\in \mathrm{Spec}(S^{-1}R)$. Indeed:

• We have $P^e \neq S^{-1}R$, because if we suppose contrary that $P^e\ = S^{-1}R$, then $P^e\ = S^{-1}R\implies (P^e)^c\ = (S^{-1}R)^c \iff P^{ec}\ = (S^{-1}R)^c \implies P = (S^{-1}R)^c$ (and the last implication becomes from a proposition in which under this hypothesis $P^{ec}=P$). But, we observe that \begin{alignat*}{2} (S^{-1}R)^c \quad = \quad &\nu^{-1}(S^{-1}R) \\ \quad = \quad & \{ x\in R: \nu(x)\in S^{-1}R \} \\ \quad = \quad & \Big\{ x\in R: \frac{x}{1_R} \in S^{-1}R \Big\} \\ \quad = \quad & R. \end{alignat*} Thus, $P=(S^{-1}R)\iff P=R$, conctradiction, because from hypothesis $P \in \mathrm{Spec}(R)$.

• Let's take two elements $\alpha:=\frac{r}{s}\in S^{-1}R$ and $\alpha':=\frac{r'}{s'}\in S^{-1}R$ s.t. $\alpha \alpha'\in P^e$. Then, from the well known proposition we take \begin{alignat*}{2}\alpha \alpha'=\frac{r}{s}\cdot \frac{r'}{s'}=\frac{rr'}{ss'}\in P^e \implies & rr' \in P \implies r\in P \text{ or } r'\in P \\ \implies & \frac{r}{s}\in P^e \text{ or } \frac{r'}{s'}\in P^e \\ \iff & \alpha \in P^e \text{ or } \alpha'\in P^e, \end{alignat*} so we get $P^e\in \mathrm{Spec}(P^{-1}R)$, as we wanted.

b) Let $P'\in \mathrm{Spec}(S^{-1}R)$ a prime ideal of $S^{-1}R$.

1) We will show that $(P')^c \in \mathrm{Spec}(R)$. One can easily prove that a contraction of a prime ideal is again a prime ideal. So, $P'\in \mathrm{Spec}(S^{-1}R) \implies (P')^c \in \mathrm{Spec}(R)$. $\checkmark$

2) Now, we will show that $(P')^c\cap S=\emptyset$.

First of all, it is a well known lemma which states that $\mathcal{ID}(S^{-1}R)=\mathcal{E}(S^{-1}R) $. So, for the ideal$P'\trianglelefteq S^{-1}R$ we conclude that $$P' \in \mathcal{ID}(S^{-1}R)=\mathcal{E}(S^{-1}R) \iff \exists I \trianglelefteq R:\ P'=I^e.$$ Then, it is $$ (P')^c=I^{ec} \iff (P')^{ce}=I^{ece} \iff (P')^{ce}=I^{e} \iff (P')^{ce}=P'.$$

We contrary suppose that $(P')^c\cap S \neq \emptyset$.

Then, there is some $x \in (P')^c\cap S \iff x\in (P')^c$ and $x\in S$.

We observe that the element $x\in (P')^c\trianglelefteq R$ and simultaneously $x\in S$. Sot the element $\frac{x}{x}\in ((P')^c)^e=(P')^{ce}$ and from this we take $$1_{S^{-1}R}=\frac{1_R}{1_R}=\frac{x}{x} \in (P')^{ce} = P' \iff P'=S^{-1}R.$$ Contradiction! Because we supposed that $P'\in \mathrm{Spec}(S^{-1}R)$.

Finally, $(P')^c\cap S = \emptyset$. $\checkmark$

Questions.

1) Could you please verify me if all I wrote above is completely correct?

2) Could you please write me down step by step number II. ?

Thank you!

Answer 1

Be careful: it is true that $P^e=\{x/s:x\in P,s\in S\}$, but it's not automatic that $x/s\in P^e$ implies $x\in P$, as you seem to be using. This is true for prime ideals not intersecting $S$. Indeed, if $$ \frac{x}{s}=\frac{y}{t} $$ with $y\in P$, then $u(tx-sy)=0$, for some $u\in S$; therefore $tux=suy\in P$; since $tu\notin P$ (as $P\cap S=\emptyset$), we conclude $x\in P$. But notice that the statement doesn't generally hold for arbitrary ideals of $R$.

Let's state it officially.

Lemma. If $P$ is a prime ideal of $R$ and $P\cap S=\emptyset$, then $x/s\in P^e$ if and only if $x\in P$.

With the above lemma, we can shorten the proofs.

Proof of Ia

Let $P$ be a prime ideal of $R$, with $P\cap S=\emptyset$. If $P^e=S^{-1}R$, then $1/1\in P^e$, hence $1\in P$: contradiction.

Suppose now that $(x/s)(y/t)\in P^e$; then $(xy)/(st)\in P^e$, so $xy\in P$.

Proof of Ib

If $f\colon A\to B$ is a ring homomorphism and $P$ is a prime ideal of $B$, then $f^{-1}(P)$ is a prime ideal of $A$. This follows immediately from the definitions. In particular, if $P'$ is a prime ideal of $S^{-1}R$, then $(P')^c=\nu^{-1}(P')$ is a prime ideal of $R$. If $s\in (P')^c\cap S$, then $s/1\in P'$ by definition, so $P'$ contains an invertible element: contradiction.

Proof of II

Consider a prime ideal $Q$ of $S^{-1}R$ and set $P=Q^c$. Let's show that $P^e=Q$. Suppose $x/s\in Q$. Then also $$ \nu(x)=\frac{x}{1}=\frac{x}{s}\frac{s}{1}\in Q $$ Therefore $x\in P=Q^c$ and therefore $x/s\in P^e$.

Suppose $x\in P$ and take any $s\in S$. Then $\nu(x)=x/1\in Q$ and so $$ \frac{x}{s}=\frac{x}{1}\frac{1}{s}\in Q $$ Conversely, if $Q=P^e$, for some prime ideal $P$ of $R$, then $P=Q^c$ (easy proof).