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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 $\mathbb{Q}$

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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 $\frac{n}{p^e}$.

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  • $\begingroup$ I guess I'm not sure how to justify 0 in S then. Are you referring to 0 as the additive identity or from inverses? $\endgroup$ – user551155 Mar 16 '20 at 16:00
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    $\begingroup$ 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. $\endgroup$ – user759746 Mar 16 '20 at 16:17
  • $\begingroup$ Ahhhhh, makes sense! Thanks so much! $\endgroup$ – user551155 Mar 16 '20 at 16:22
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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 $\mathbb Q$.

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