I am working in Adkins & Weintraub, and they ask the reader to consider the integers adjoined with a fraction $\dfrac{p}{q}$. I'm not really sure what they mean by this, since they didn't mention this construction in the chapter on rings(Or at least, I didn't $\textit{notice}$ any such construction...). Do they mean the localization away from some multiplicatively closed subset of the integers so that $\dfrac{p}{q}$ becomes invertible?
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$\begingroup$In general, given a commutative ring $R$ and an element $\alpha$ lying in some ring $E$ where $R \subseteq E$ we have the subring $\langle R \cup \{\alpha\}\rangle = \{f(\alpha): f \in R[x]\}$.
This is forced upon us by the closure axioms of a ring, polynomial rings over a commutative ring can be thought of as "universal adjuction rings". The above statement is usually framed this way:
There is a unique surjective ring-homomorphism $R[x] \to R[\alpha]$ that sends $x \to \alpha$ (sometimes this is phrased in terms of a similar homomorphism $R[x] \to E$). The astute reader will note this defines a universal mapping property which essentially turns the rings $R[x]$ into "free objects" of some kind.
One can think of this as "polynomial expressions in $\alpha$", but beware: it can happen that $f(\alpha) = 0$ in $E$, so these expressions aren't always unique. For example, when $R = \Bbb Z$, and $\alpha = \frac{p}{q} \in \Bbb Q$ with $\gcd(p,q) = 1$ then we can write $k = rp + sq \in \Bbb Z$, and the polynomials expressions induced by $f(x) = k$ and $g(x) = rqx + sq$ are, in fact, the same element of $\Bbb Q$.
This is one of the reasons why polynomial rings are so important, they serve as templates for understanding (simple) extension rings.
$\endgroup$ $\begingroup$Any time you adjoin an element $\alpha$ to a ring $R$, you adjoin all elements of the form $a + b \alpha$ such that $a, b \in R$. So in this case, taking $\mathbb{Z}\left[\frac{p}{q}\right]$, you include all elements of the form $f\left( \frac{p}{q} \right)$, where $f(x) \in \mathbb{Z}[x]$.
I believe this is the aforementioned construction. As @dorebell said, this ring is equivalent to $\mathbb{Z}[x]/(xq-p)$. Any time you adjoin an element somehow "outside" a ring, you tend to make it out of some polynomial with the ring coefficients.
EDIT: Further, note if $f$ is an integer polynomial and $q$ divides all the coefficients of $f$, then $f\left( \frac{p}{q} \right)$ is in $\mathbb{Z}$. As @Travis mentioned in the comments, the most notable example is $qx-p$, when evaluated at $\frac{p}{q}$ is zero. This is the minimal polynomial for $\frac{p}{q}$ in $\mathbb{Z}$, so if $f\left( \frac{p}{q} \right) = 0,$ then $qx-p | f$.
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