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Consider the ring $R$ of rational numbers which, when written in simplest form, have an odd denominator. This is a subring of $\mathbb{Q}$ with the usual multiplication and addition.

I wonder if you could check my understanding of the units, irreducible elements, and prime elements in this ring.

Assume all fractions are in simplest form.

Units: For any element $\frac{a}{b} \in R$ either $a$ is odd or $a$ is even. If $a$ is odd, then $\frac{b}{a} \in R$ and $\frac{a}{b}\frac{b}{a}=1$. Hence, $\frac{a}{b}$ is a unit. If $a$ is even, then $\frac{b}{a}$ is not in $R$. There's no other possibility for an inverse, so $\frac{a}{b}$ is not a unit.

Irreducible elements: We can write an element $\frac{a}{b}$ uniquely in the form $\frac{2^ne}{b}$ where $n \in \mathbb{N}$ and $e$ is odd. I claim that that the irreducible elements are precisely those for which $n=1$. Proof: If $n=0$, then $a$ is odd and our element $\frac{a}{b}$ is a unit, not irreducible. If $n>1$, then $\frac{a}{b}=\frac{2e}{1}\frac{2^{n-1}}{b}$ and neither of the elements of this product are units because their respective numerators are even. However, if $n=1$, we have

$$ \frac{a}{b}=\frac{2e}{b}=\frac{w}{v}\frac{y}{z}. $$

Then one of $w,y$ must be even and the other must be odd. Suppose WLOG that $w$ is even. Then $y$ is odd and $\frac{y}{z}$ is a unit. Hence, $\frac{a}{b}$ is irreducible.

Primes: All primes must be irreducible, so we may look among the irreducible elements for our primes. That is, among the elements of the form $\frac{2e}{b}$ with $e$ odd. I'm having trouble getting any further, though.

Any suggestions?

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2 Answers2

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As an alternative approach, what you're really doing here is taking a localization of $\mathbb{Z}$. In fact, if we let $$S=\{\pm 1\}\cup \{\pm p_1p_2 \cdots p_n\,\vert\,p_i \textrm{ is an odd prime }\}$$ then $R=\mathbb{Z}_S=\{\frac{a}{b}\,\vert\,a\in \mathbb{Z}, b\in S\}$ is the ring you're looking for.

In fact, since the prime ideals of $\mathbb{Z}_S$ are precisely the extension of the prime ideals in $\mathbb{Z}$ that intersect trivially with $S$, it's not too tough to see that, in fact, $R=\mathbb{Z}_{2\mathbb{Z}}$--ie $R$ is the localization of $\mathbb{Z}$ at the prime ideal $2\mathbb{Z}$ (which is an obfuscated way of denoting what we really mean: the localization of $\mathbb{Z}$ at $\mathbb{Z}\setminus 2\mathbb{Z}$ (for any domain $D$ and any prime ideal $P$ of $D$, $D\setminus P$ is a multiplicative subset of $D$, and by $D_P$ we really mean $D_{D\setminus P}$...the notation is horrendous, but we're stuck with it). So, the only nonzero element of $\mathbb{Z}$ that survives as a nonunit in $R$ is 2, hence any nonzero nonunit in $R$ can be written as a unit multiple of a power of 2.

Check out my answer to another, somewhat related question for a more general approach.

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You are almost done. You have shown that, up to associates, $2$ is the only possible prime in $R$. To confirm its primality you need to show that $2$ is a nonzero nonunit such that for all $r,s\in R$, if $2\nmid r,s$ then $2\nmid rs$ in $R$. This is a straightforward parity argument (lift up primality of $2$ in $\mathbb Z$).

Generally any ring that has $\mathbb Z/2$ as an image has a natural parity structure. Here one can easily verify that $R/2 \cong \mathbb Z/2$, which also yields a more "structural" verification of the primality of $2$.

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