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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.

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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.

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  • $\begingroup$ 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 $\mathbb Z\cup P^{-1}$? $\endgroup$
    – CRinge
    Commented Jul 27, 2022 at 7:57
  • $\begingroup$ @CRinge It is true, as you correctly proved in your question. $\endgroup$
    – Adayah
    Commented Aug 3, 2022 at 18:19

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