# Adjoining inverses to $\mathbb Z$

Let $$R$$ be a (commutative unital) ring. For a set $$X\subseteq R$$, and any subring $$S\subseteq R$$, Pete Clark in his Commutative Algebra defines $$S[X]$$ as the smallest (resp. to inclusion) subring of $$R$$ containing $$S$$ and $$X$$. I guess the term adjoining is used because $$S\mapsto S[X]$$ yields a functor $$f:\mathrm {Sub}(R)\to \mathrm {Sub}_X(R)$$, which is left adjoint to the inclusion $$i:\mathrm {Sub}_X(R)\to \mathrm {Sub}(R)$$: the universal property of $$f$$ is satisfied by the fact that, if $$T\subseteq R$$ is a ring containing $$X$$, and $$S\subseteq T$$, then $$S[X]\subseteq T$$. The notation $$\mathrm {Sub}(R), \mathrm {Sub}_X(R)$$ stands for, resp., the poset category of the subrings of $$R$$ and that of the subrings of $$R$$ containing $$X$$.

I'm trying exercise 1.6: show that $$\mathbb Z[P^{-1}]\cong \mathbb Z[Q^{-1}]$$, $$\mathbb Z[P^{-1}]= \mathbb Z[Q^{-1}]$$, $$P=Q$$ are equivalent, where $$P,Q$$ are any two sets of prime integers, and $$Z^{-1}:=\{\frac 1z\in \mathbb Q:z\in Z\}$$ for any $$Z\subseteq \mathbb Z$$ (with $$0\notin Z$$).

• $$\mathbb Z[P^{-1}]\cong \mathbb Z[Q^{-1}]\implies \mathbb Z[P^{-1}]= \mathbb Z[Q^{-1}]$$. Take an isomorphism $$\phi:R\to S$$ of subrings $$\mathbb Z\subseteq R,S\subseteq \mathbb Q$$, and let $$r\in R$$. Then $$mr=n$$ for some $$m,n\in \mathbb Z$$; $$\phi$$ is a homomorphism of $$\mathbb Z$$-algebras, so $$m\phi(r)=n$$, and $$r=\phi(r)$$ in $$\mathbb Q$$.
• The other serious implication is $$\mathbb Z[P^{-1}]= \mathbb Z[Q^{-1}]\implies P=Q$$. I'd prove that $$P\mapsto \mathbb Z[P^{-1}]$$ is an injective map from the sets of prime integers to the intermediate rings between $$\mathbb Z$$ and $$\mathbb Q$$. A left inverse is $$R\mapsto \{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}$$. Then $$P\subseteq \{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in \mathbb Z[P^{-1}] \}$$; but how to prove that, for a prime integer $$q\notin P$$, $$\frac 1q \notin \mathbb Z[P^{-1}]$$? The author avoided defining $$\mathbb Z[P^{-1}]$$ explicitly, so I wander if one can prove it just by the universal property that $$\mathbb Z[P^{-1}]$$ is the smallest subring of $$\mathbb Q$$ containing $$P^{-1}$$.

It may help exercise 1.7, that asks to prove also that $$R\mapsto \{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}$$ is a right inverse to $$P\mapsto \mathbb Z[P^{-1}]$$, i.e. $$\mathbb Z[\{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}^{-1}]=R$$. Since $$\{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}^{-1} \subseteq R$$, by the universal property $$\mathbb Z[\{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}^{-1}]\subseteq R$$. Conversely, take $$\frac mn\in \mathbb Q$$: one has $$\frac mn\in R$$ iff $$\frac 1n\in R$$ (by Chinese RT) iff $$\frac 1p\in R$$ for all primes $$p$$ that divide $$n$$. Any $$r\in R$$ is the product in $$\mathbb Q$$ of elements in $$\mathbb Z\cup \{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}^{-1}$$ then, and $$r\in \mathbb Z[\{p\in \mathbb Z:p\mathrm {\ is\ prime\ and\ }\frac 1p\in R \}^{-1}]$$. In this way, one can also prove what was left from the second point, because if $$q\notin P$$, clearly $$\frac 1q$$ cannot be obtained multiplying elements of $$\mathbb Z$$ and $$P^{-1}$$. But since this exercise comes after 1.6, maybe there is a more immediate solution to the second point, using the universal property.

But how to prove that, for a prime integer $$q \notin P$$, $$\frac{1}{q} \notin \mathbb{Z}[P^{-1}]$$?

The set

$$R = \left\{ \frac{a}{b} \in \mathbb{Q} : q \nmid b \right\}$$

is a subring of $$\mathbb{Q}$$ containing $$\mathbb{Z} \cup P^{-1}$$, so $$\mathbb{Z}[P^{-1}] \subseteq R$$ and so $$\frac{1}{q} \notin \mathbb{Z}[P^{-1}]$$.

Also

In this way, one can also prove what was left from the second point, because if $$q \notin P$$, clearly $$\frac{1}{q}$$ cannot be obtained multiplying elements of $$\mathbb{Z}$$ and $$P^{-1}$$.

is not enough - you would have to prove that $$\frac{1}{q}$$ can not be expressed as a sum of such products.

• Thanks, I understand the first part. Isn't true, though, that if $\mathbb Z\subseteq R\subseteq\mathbb Q$ is a ring, and $P:=\{p\in \mathbb Z:p\mathrm{\ is\ prime\ and\ }\frac 1p\in R\}$, every $r\in R$ can be written in $\mathbb Q$ as the product of elements in $\mathbb Z\cup P^{-1}$? Commented Jul 27, 2022 at 7:57
• @CRinge It is true, as you correctly proved in your question. Commented Aug 3, 2022 at 18:19