Rings of fractions for all possible sets of denominators

This is related to a question I asked here except in this case, I am not asking about the total (complete) ring of fractions. I am asking how to find all possible rings of fractions up to isomorphism for all possible denominators.

Specifically, I am asking for the cases when the rings in question are

• $R = \mathbb{Z}_{p}$, where $p$ is a prime. (Note: This is NOT the p-adic integers. This is the set $\{0, 1, \cdots, p-1 \}$ of integers modulo a prime $p$.)
• $R=\mathbb{Z}_{p^{2}}$, again, where $p$ is a prime.

I know that when $R = \mathbb{Z}_{p}$, which is a field, the only multiplicative subgroups are the zero subgroup, $U(\mathbb{Z}_{p})$ (the units of $\mathbb{Z}_{p}$), and of course $\mathbb{Z}_{p}$ itself (although I don't know how to prove this).

When we are discussing the total ring of fractions, the set $D$ (sometimes called $S$) of denominators consists of all non-zero, non zero divisor elements of $\mathbb{Z}_{p}$. However, here, I am asked to consider ALL possible denominators. When $D$ contains $0$, what I get for my ring of fractions is the zero ring, correct (so this would also include the case when $D$ is the whole ring?)? When $D$ is the set of units, then I have the total ring of fractions, and I have, by the question I linked, that $D^{-1}\mathbb{Z}_{p} \simeq \mathbb{Z}_{p}$? Is any of what I'm reasoning here correct?

But what about thee case when $R = \mathbb{Z}_{p^{2}}$? What are all the possible sets of denominators in this case? Recall that the set of denominators just has to be a sub-semigroup. So maybe even for the case where $\mathbb{Z}_{p}$, I'm not including all the possible sets of denominators!

Where I'm getting stuck is - I don't know how to decide what all the $D$'s should be that I need to check. Because inversion by units doesn't change anything from the complete (total) ring of fractions, I suppose I should just consider $D$'s containing zero divisors and a $D$ consisting of just $0$ itself? Then, how do I figure out what they're isomorphic to?

• @BillDubuque can you explain more? I need to consider all possible sets of divisors. So how do I know that I've considered them all? What are they? – ALannister Jul 8 '17 at 22:06
• You know that inverting units does nothing. So it remains to consider what happens when you invert the zero-divisors like $\,ap.\$ – Bill Dubuque Jul 8 '17 at 22:09
• @BillDubuque do you get the zero ring when you do that? Because isn't that the case when 1=0, and so you have to have the zero ring? – ALannister Jul 8 '17 at 22:12
• Post your argument as an answer and we will give you feedback (or elaborate in your question where you get stuck). – Bill Dubuque Jul 8 '17 at 22:15
• $U(p)$ is a cyclic group of order $p - 1$. For example with $p = 5$ the group is isomorphic go $Z_4$ which has more than two subgroups. For instance $\{1,4\}$ is a subgroup of $\mathbf{Z}/5$ with two elements. – Trevor Gunn Jul 8 '17 at 22:17

• $R = \mathbb{Z}_{p}$, $p$ prime:

First, note that since $p$ is prime, $\mathbb{Z}_{p}$ is a field; i.e., $\mathbb{Z}_{p}$ has no zero divisors. I.e., $\forall m \in \mathbb{Z}_{p}$ where $m \neq 0$, $\exists n \in \mathbb{Z}_{p}$such that $mn = nm = 1$ -i.e., every nonzero element of $\mathbb{Z}_{p}$ is a unit.

Now, suppose $D$ consists only of units, then $D \subseteq U( \mathbb{Z}_{p})$. Also, note that $\mathbb{Z}_{p}$ has the identity. So, we apply the following result:

Proposition - Let $R$ be a commutative ring and $D$ be a nonempty multiplicative subsemigroup of $R$. Then the following is true:

1. The map $\phi: R \to D^{-1}R$ defined by $$\phi(x) = \frac{xy}{y}\, \text{for some}\,y \in D.$$ is a ring homomorphism.
2. If $0 \notin D$ and $D$ has no zero divisors, then $\phi$ is a monomorphism.
3. The image $\phi(x)$ of every element $x \in D$ is invertible; i.e., $\phi (D) \subseteq U(D^{-1}R)$.
4. If $R$ has the identity and $D \subseteq U(R)$ then $\phi$ is an isomorphism.

to conclude that the map $\displaystyle \phi: \mathbb{Z}_{p}\to D^{-1}\mathbb{Z}_{p}$ defined by $\displaystyle \phi(x) = \frac{xy}{y}$ for some $y \in D$ is an isomorphism. Thus, in the case when $D$ consists only of units, the ring of fractions is $D^{-1}\mathbb{Z}_{p} \simeq \mathbb{Z}_{p}$.

Second, suppose instead that $0 \in D$. In order for the pairs $(x,y)$, $(x^{\prime},y^{\prime}) \in R \times D$ to be equal, there must exist some $z \in D$ such that $$(xy^{\prime}-yx^{\prime})z = 0.$$

Now, according to my notes, this guarantees that $D^{-1}R$ must be the zero ring, but I am not sure why in terms of $(xy^{\prime} - yx^{\prime})z=0$ that is true. If someone could explain that to me either in the comments or in a different answer, I would be most appreciative.

So, to summarize, in the case that $0 \in D$, $D^{-1}\mathbb{Z}_{p} = \{0\}$.

• $R = \mathbb{Z}_{p^{2}}$, $p$ prime:

In the case that $R = \mathbb{Z}_{p^{2}}$, the set of units is the set of all $m \in \mathbb{Z}_{p^{2}}$ such that $\gcd(m, p^{2}) = 1$.

So, suppose first that $D$ consists only of units of $\mathbb{Z}_{p^{2}}$. Then, $D \subseteq U(\mathbb{Z}_{p^{2}})$, and we note that $\mathbb{Z}_{p^{2}}$ has the identity. Therefore, by the proposition given in the first bullet point, the map $\phi: \mathbb{Z}_{p^{2}} \to D^{-1}\mathbb{Z}_{p^{2}}$ defined by $\displaystyle \phi(x)=\frac{xy}{y}$ for some $y \in D$ is an isomorphism. Thus, in the case when $D$ consists only of units, the ring of fractions is $D^{-1}\mathbb{Z}_{p^{2}} \simeq \mathbb{Z}_{p^{2}}$.

Second, suppose that $0 \in D$. By the remarks in the first bullet point, again, this is the case where $D^{-1}\mathbb{Z}_{p^{2}}$ is the zero ring.

Third, suppose that $D$ contains a zero divisor. Now, zero divisors of $\mathbb{Z}_{p^{2}}$ are multiples of $p$.

Since $D$ is a multiplicative semisubgroup, if $ap \in D$ where $ap$ is a multiple of $p$, then $D$ contains $(ap)\cdot (ap) = (ap)^{2} = a^{2}p^{2} = a^{2} \cdot 0 = 0$.

So, if $D$ contains a zero divisor, it also contains $0$ itself, which degenerates to the case where $0 \in D$; i.e., in this case also, $D^{-1}\mathbb{Z}_{p^{2}}$ is the zero ring.

• I will use slightly different letters to make clear which elements belong to $D$. We have $\frac{x_1}{s_1} = \frac{x_2}{s_2}$ if and only if there exists $r \in D$ such that $r(s_2x_1 - s_1x_2) = 0$. If I am allowed to take $r = 0$ then I have $0(s_2x_1 - s_1x_2) = 0$. This is always true regardless of what $x_1,x_2,s_1,s_2$ are. Thus all fractions in $D^{-1}R$ are equal. If you have a ring where every two elements are equal then that ring is the zero ring. – Trevor Gunn Jul 9 '17 at 21:30
• The rest of your answer reflects a good understanding of the problem. – Trevor Gunn Jul 9 '17 at 21:45
• @T.Gunn thank you for everything! – ALannister Jul 9 '17 at 23:16