Take $R = \mathbb{Z}$ and fix a set of nonzero integers $X = \{n_{1},n_{2},\dots, n_{k} \}$. Now, let $$D = \{n_{1}^{t_{1}}n_{2}^{t_{2}}\cdots n_{k}^{t_{k}} \mid t_{i}\geq 0 \, \text{for all}\, i,\ t_{i}\neq 0\, \text{for at least one}\, i\}$$ i.e., $D$ is a semisubgroup generated by $X$.

I need to show that $D^{-1}\mathbb{Z}$ is isomorphic to another ring of fractions $E^{-1}\mathbb{Z}$ where $E$ is generated by a single number $n$ - i.e., $$E = \{n^{t}\mid t>0\} $$

but I am not sure how to go about doing this.

What I do know is that for $m \in \mathbb{Z}$, elements of $D^{-1}\mathbb{Z}$ are of the form

$$ \frac{m}{c\cdot n_{1}^{t_{1}}\cdot n_{2}^{t_{2}} \cdots n_{k}^{t_{k}}}, \, \text{where}\,c\,\text{is a constant} $$

Now, I assume that elements of $E^{-1}\mathbb{Z}$, where $m^{\prime} \in \mathbb{Z}$ are of the form $$ \frac{m^{\prime}}{a\cdot n^{t}}, \, \text{where}\, a\, \text{is a constant}.$$

Therefore, I imagine that what I need to do is find some kind of isomorphism between these two sets of elements.

Should I use the universal properties of rings of fractions for this? If so, which ones? Is there a way I can say that $D^{-1}\mathbb{Z}$ is embedded in $E^{-1}\mathbb{Z}$? Maybe by letting $\displaystyle n^{t}:=\prod_{i=1}^{k}n_{i}^{t_{i}}$?

Or should I try to embed each of them into $\mathbb{Q}$, since each ring of fractions can be thought of as a subring of $\mathbb{Q}$?

Any help you could give me would be much appreciated! Thank you.

  • $\begingroup$ Assuming $D^{-1} \mathbb{Z} = E^{-1} \mathbb{Z}$, what happens when you add one element to $D$ ? $\endgroup$ – reuns Jul 10 '17 at 2:02
  • $\begingroup$ Two comments. First, it doesn't matter if you let $t_1 = \dots = t_k = 0$ because then you have $n_1^0 \cdots n_k^0 = 1$ and having $\frac{1}{1}$ as a fraction is OK. Second, there is no extra constant in the denominator: $$D^{-1}\mathbf{Z} = \left\{ \frac{m}{n_1^{t_1} \cdots n_k^{t_k}} : m \in \mathbf{Z}, t_i \ge 0 \right\}.$$ $\endgroup$ – Trevor Gunn Jul 10 '17 at 2:07
  • $\begingroup$ Hint $\ a,b,c$ are units $\iff abc$ is a unit, i.e. units form a saturated monoid $\, u,v\in U \iff uv\in U\ \ \ $ $\endgroup$ – Bill Dubuque Jul 11 '17 at 2:08

Suppose you are given a ring $R$ and an element $n \in R$ with $n^k \ne 0$ for any $k \ge 0$. Let $D = \{1, n, n^2, \dots\}$. Then

$$ D^{-1}R = \left\{ \frac{x}{n^k} : k \ge 0 \right \} $$

which we can view as taking the ring $R$ and adding an inverse $\frac1n$ for $n$.

It is exactly the same situation when we have the subsemigroup generated by $\{n_1,\dots,n_k\}$, namely, we are adding inverses to $n_1,\dots,n_k$. Since $R = \mathbf{Z}$ all fraction rings of $\mathbf{Z}$ are subrings of $\mathbf{Q}$. In this case we are looking at the subring containing $\mathbf{Z}$ and each inverse $\frac{1}{n_i}$ for $i \in \{1,\dots,k\}$.

For example, if you let $D$ be generated by $2$ and $3$ then you are looking at all rational numbers of the form

$$ \frac{x}{2^a3^b}. $$

Can you see why every such fraction can be written in the form $\dfrac{y}{6^c}$? Spoiler:

$$ \dfrac{x}{2^a3^b} = \dfrac{x2^b3^a}{2^a2^b3^a3^b} = \dfrac{x2^b3^a}{6^{a + b}}. $$

Given this, a reasonable common denominator for $n_1,\dots,n_k$ would be $n = n_1 \cdots n_k$.

Using what we've seen for $\{2, 3\}$ we will try to write the fraction

$$ \frac{x}{n_1^{a_1}\cdots n_k^{a_k}} \text{ as } \frac{y}{n^b}. $$


Let $b = a_1 + \dots + a_k$ and let $y = xn_1^{b - a_1} \cdots n_k^{b - a_k}$$.

Thus $D^{-1}\mathbf{Z} = E^{-1}\mathbf{Z}$ where $n = n_1\cdots n_k$.


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