# Show that S is a subring of $\mathbb{Q}$

Can somebody please look at my work and critique. Thanks in advance!

Let p be a prime number, let $$\mathbb{Q}$$ be the field of rational numbers, and define the set: $$S = \{{n/p^e \,|\, n \in \mathbb{Z}, e \in \mathbb{Z}}\} \subset \mathbb{Q}$$ Show that S is a subring of $$\mathbb{Q}$$.

First we show that ($$S,+$$) is a subgroup of ($$\mathbb{Q},+$$):

$$n/p^e + n/p^e = 2n/p^e \in S \qquad$$ closed under addition

$$n/p^e + 0 = n/p^e \in S \qquad$$additive identity

$$n/p^e + \left(-n/p^e\right) = 0 \in S \qquad$$ Inverse

Hence, ($$S,+$$) is a subgroup of ($$\mathbb{Q},+$$)

Next we show ($$S,\times$$) is closed and $$1 \in S$$:

$$\left(n/p^e\right) \left(n/p^e\right) \in S \qquad$$ closed under multiplication

$$\left(n/p^e\right) \left(n/p^e\right)^{-1} = 1 \in S$$

Hence, S is a subring of $$\mathbb{Q}$$

I think only $$p$$ is fixed here. To show that it is closed under addition, you should show that $$\frac{m}{p^e}+\frac{n}{p^f}=\frac{p^fm+p^en}{p^{e+f}}\in S$$, for example. That $$0$$ would be the additive identity is inherited from $$\mathbb{Q}$$, but you give no justification for $$0\in S$$. You don't prove that the set $$S$$ is closed under products, you only state it is. To fix this, you should give some reason for $$\frac{m}{p^e}\cdot\frac{n}{p^f}=\frac{mn}{p^{e+f}}$$ to be in $$S$$. Also, your proof for $$1\in S$$ contains the risk that you divide by $$0$$, and it can be bypassed simply by showing that you can write $$1$$ in the form $$\frac{n}{p^e}$$.
• You can show that $0\in S$ simply by choosing $n=0$ in your definition, but you can also do it by showing first that $S$ is additively closed, and that if $\frac{n}{p^e}\in S$ then $-\frac{n}{p^e}\in S$, which will take care of the additive inverses at the same time. – user759746 Mar 16 '20 at 16:17
Alternatively, consider the map $$\mathbb Z[x] \to \mathbb Q$$ induced by $$x \mapsto \frac 1p$$.
This map is a ring homomorphism whose image is $$S$$. Therefore, $$S$$ is a subring of $$\mathbb Q$$.