Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Algebra / 3

Interpret membership as having a representation with denominator coprime to $30$, equivalently having this property in lowest terms. Clearly $0,1\in R$. If $a/b,c/d\in R$ with admissible denominators, thenThe product $bd$ is still coprime to $30$. These are elements of $R$, proving that $\boxed{R\text{ is a subring of }\mathbb{Q}}$.

Take a reduced fraction $a/b\in R$. If $a\ne 0$, its only possible inverse in $\mathbb{Q}$ is $b/a$. Since $\gcd (a,b)=1$, this is in lowest terms up to sign and belongs to $R$ exactly when $\gcd (a,30)=1$. Zero is not invertible. ThereforeEquivalently, these are the rational numbers whose valuations at $2,3,5$ are all zero.

The intersection $J=I\cap \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$, and $nj\in I\cap \mathbb{Z}$ for $n\in \mathbb{Z}$, $j\in J$. Thus it is an ideal of $\mathbb{Z}$, necessarily $d\mathbb{Z}$ for a unique $d\ge 0$.

Because $d\in I$, one has $dR\subseteq I$. Conversely, if $a/b\in I$ with $b$ coprime to $30$, multiplication by $b\in R$ gives $a\in I\cap \mathbb{Z}=d\mathbb{Z}$. Write $a=dk$; then $a/b=d(k/b)\in dR$. Thus $I=dR$.

Every positive integer factors as $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }u$ with $\gcd (u,30)=1$. The factor $u$ is a unit, so it does not change the ideal. The complete distinct list isThey are distinct because the intersection of the displayed nonzero ideal with $\mathbb{Z}$ is $2^{\alpha }{3}^{\beta }{5}^{\gamma }\mathbb{Z}$: admissible denominators cannot cancel any of these prime powers. The triple $(0,0,0)$ gives $R$ itself.

The zero ideal is a prime ideal because $R$ is an integral domain. Each of $(2),(3),(5)$ is a prime ideal: for $p$ among these three, reduction $a/b\mapsto \overline{a}{\overline{b}}^{-1}$ is a surjective ring homomorphism $R\to {\mathbb{F}}_p$ with kernel of a ring homomorphism $(p)$, so its quotient ring is a field.

Any other nonzero proper ideal in (c) has generator $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ with total exponent at least two. Choose one prime factor $p$ and put $e=d/p$. Then $pe=d\in (d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required prime exponents. Such an ideal is not prime. Thus with the standard proper-ideal convention the answer isThe definition printed in this paper omits properness. Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four ideals together with $\boxed{R}$. Both interpretations are thereby accounted for.