3

A conventional wisdom of building $\mathbb{Q}$ from $\mathbb{Z}$ is the construction of the Field of Quotients of the Integral Domain. Are there alternatives?

For example, can the $\mathbb{Q}$ be described as a Quotient Ring: $\mathbb{Z}[x,y]/\langle xy-1 \rangle$? I struggle to exhibit an isomorphism of this ring into $\mathbb{Q}$.

Reflecting onto the standard construction of the Field of Quotients

$\mathbb{Z} \oplus \mathbb{Z} / \{ (a,b) \sim (c,d) \mid ad = bc \} $

it prompts that perhaps we need to construct a projective variety out of $\mathbb{Z} \oplus \mathbb{Z}$ (if there is such a thing), and work with homogeneous polynomials?

In general, given the 2 ring building operations: direct sum, and making a polynomial ring, is there a ring $R(\mathbb{Z})$, e.g.

$ (\mathbb{Z} \oplus \mathbb{Z})[x,y,z] \oplus (\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z})[x,t] $

and an ideal $I \subset R(\mathbb{Z})$ such that $ R(\mathbb{Z})/I$ is isomorphic to $\mathbb{Q}$?

Noa Even
  • 2,801
Tegiri Nenashi
  • 1,030
  • 2
    $\mathbb{Z}[x,y,z]/\langle xy-z \rangle$ is simply isomorphic to $\mathbb{Z}[x,y]$; indeed this is true of any $\mathbb{Z}[x,y,z]/\langle p(x,y)-z \rangle$ where $p$ is a polynomial (consider $f\colon \mathbb{Z}[x,y,z]\to\mathbb{Z}[x,y]$ given by $f(x)=x$, $f(y)=y$, $f(z) = p(x,y)$). – Greg Martin Dec 08 '21 at 18:25
  • How about $\mathbb{Z}[x,y]/\langle xy-1 \rangle$ – Tegiri Nenashi Dec 08 '21 at 18:29
  • 1
    @TegiriNenashi No. If there was a isomorphism $f:\mathbb Z[x,y]/\langle xy-1\rangle\to\mathbb Q$, then $f$ is the identity on $\mathbb Z$, $f(x)\neq0$, and $f(y)=\frac1{f(x)}$. Thus, $\frac{f(x)}2$ is not mapped to. – Rushabh Mehta Dec 08 '21 at 18:36
  • 2
    $\mathbb{Z}[x,y]/\langle xy-1\rangle$ is just isomorphic to $\mathbb{Z}[t,t^{-1}]$. You did not even add an inverse to $2$, so how could it be isomorphic to $\mathbb{Q}$? – Arturo Magidin Dec 08 '21 at 18:38
  • You can obtain it as a quotient by adding countably many independent variables indexed by primes, say $x_p$, $p$ a prime, and moding out by the ideals generated by all $px_p-1$. – Arturo Magidin Dec 08 '21 at 18:39
  • 2
    $\mathbb{Q}\cong \mathbb{Z}[{x_p\mid p\gt 0,p\text{ a prime}}]/\langle px_p-1\mid p\gt 0, p\text{ a prime}\rangle$. – Arturo Magidin Dec 08 '21 at 18:49
  • 1
    Your ring is finitely generated as a $\Bbb Z$-algebra, but $\Bbb Q$ is not. – Lubin Dec 08 '21 at 22:55

2 Answers2

2

Yes, indeed, there are alternatives to the common pair-based (normal form) construction of rings of fractions. Below are a couple alternatives.

The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way, and see this answer for a simplification to both expositions, and generalizations (e.g. to noncommutative rings).

Note that pairs $(a,s)$ representing $a/s$ in the common pair-based approach are simply convenient "normal" forms of the polynomials used in the above presentation based approach (with the ring structure transported to the normal forms in standard way, e.g. see here).

Another approach - not as well-known - is via partial endomorphisms - see below, excerpted from Lambek's book Lectures on Rings and Modules.

enter image description here enter image description here enter image description here

Bill Dubuque
  • 272,048
2

This is essentially the same as Bill Dubuque’s first part, but note that it suffices to find an epimorphism from a constructible (by the allowed operations) ring to the rationals. With a polynomial ring $\Bbb Z[X_1,X_2,X_3,\ldots]$ in infinitely many $X_i$, we can achieve this by mapping $X_i\mapsto \frac 1i$. Remains the task of explicitly describing the kernel, but at least it is immediate that this works out.

What if we allow only finitely many variables and finitely many $\oplus$? In that case, all is determined by where the finitely many variables and finitely many ones are mapped. This cannot be onto, e.g., we will miss $\frac1p$ where $p$ is a large prime not “used” by any of those finitely many values.

Hagen von Eitzen
  • 374,180