Show that $\mathbb Z_{(p)} / J(\mathbb Z_{(p)}) \cong \mathbb Z_p$

Question

For some context, this is one of the exercises on a section about ring homomorphisms. One of the definitions in this exercise is of a local ring (I googled it and it looks like it has a different definition and they may or may not be equivalent):

A ring $R$ is a local ring if the set $J(R)$ of nonunits in $R$ forms an ideal.

Some of the relevant properties of a local ring are:

1. Every division $R$ is local.
2. If $p$ is prime, then $\mathbb Z_{(p)} = \{ \frac{n}{m} \in \mathbb Q \mid p \text{ does not divide } m \}$ is local.
3. If $R$ is local, then $R/J(R)$ is a division ring.

Now the exercise (Exercise 3.4.28 Introduction to Abstract Algebra by Nicholson) asks:

If $p$ is a prime, let $\mathbb Z_{(p)} = \{ \frac{n}{m} \in \mathbb Q \mid p \text{ does not divide } m \}$ and consider the set $J(\mathbb Z_{(p)}) = \{ \frac{n}{m} \in \mathbb Z_p \mid p \text{ divides } n\}$. Show that $J(\mathbb Z_{(p)})$ is an ideal of $\mathbb Z_{(p)}$ and $\mathbb Z_{(p)} / J(\mathbb Z_{(p)}) \cong \mathbb Z_{(p)}$. (Note from me: the field on the right is a cyclic group, NOT the same one as before.)

I've tried applying the isomorphism theorem with the rings $\mathbb Z_{(p)}$ and $\mathbb Z_p$ for a good while but no luck, but I tried using another theorem, in particular its corollary:

Every field $R$ contains a central subfield isomorphic to $\mathbb Z_p$, or $\mathbb Q$ according as $\text{char} \; R = p$ or $\text{char} \; R = 0$.

Since $\mathbb Z_{(p)}$ is a local ring we can apply (3) here to find out that $\mathbb Z_{(p)} / J(\mathbb Z_{(p)})$ is a division ring. It's also commutative because $\mathbb Z_{(p)}$ is and therefore it is a field. Its unity is $1 + J(\mathbb Z_{(p)})$, and if we add it $p$ times we get $$ p(1 + J(\mathbb Z_{(p)})) = p + J(\mathbb Z_{(p)}) $$ but since $p = \frac{p}{1} \in J(\mathbb Z_{(p)})$ it follows that $p + J(\mathbb Z_{(p)}) = 0 + J(\mathbb Z_{(p)})$, and since $p$ is the lowest integer that does this we have $\text{char} \; J(\mathbb Z_{(p)}) = p$, so the corollary applies. Thus there exists a central subfield in $ \mathbb Z_{(p)}/J(\mathbb Z_{(p)})$ that is isomorphic to $\mathbb Z_p$, so at this point it suffices to show that this central subfield is the entire field. The proof for the corollary's theorem showed that the central subfield is $\mathbb Z 1_R = \{ k1_R \mid k \in \mathbb Z \}$, but I don't know how to apply this information because I am so tired at this point.

Another thing I noticed is that if we have a mapping $\theta: \mathbb Z_{(p)} / J(\mathbb Z_{(p)}) \to \mathbb Z_p$ defined by $\theta(\frac{m}{n} + J(\mathbb Z_{(p)})) = \text{round}(\mathbb Z_{(p)})$ that this could be an isomorphism but my brain is so fried I will not even bother to check it in fact I doubt it even is my brain is hallucinating stuff at this point.

Could anyone help me by completing this or by suggesting a better way of doing this?