# Finding the residue field for the local ring $\mathbb{Z}_{\langle 5\rangle}$ with maximal ideal $5\mathbb{Z}_{\langle 5\rangle}$

I wish to find the residue field for the local ring $$\mathbb{Z}_{\langle 5\rangle}$$ with maximal ideal $$5\mathbb{Z}_{\langle 5\rangle}$$, i.e. compute the quotient $$\mathbb{Z}_{\langle 5\rangle}/5\mathbb{Z}_{\langle 5\rangle}$$. It seems intuitively that this should be isomorphic to $$\mathbb{Z}/5\mathbb{Z}= \mathbb{F}_5$$, but I have no idea how to construct the isomorphism.

I tried by considering a map $$\mathbb{Z}_{\langle 5\rangle}\rightarrow \mathbb{Z}/5\mathbb{Z}$$, with a view to applying the First Isomorphism Theorem, which takes elements $$\frac{a}{b}$$ to $$a$$ mod $$5$$. However, I was struggling to show that this map is well-defined. Any help would be greatly appreciated!

$$5\Bbb Z_5$$ is the zero ideal. Thus we do get back the original $$\Bbb Z_5$$.
• I'm sorry but I don't understand how this is true. Clearly $\frac{5}{1}\in 5\mathbb{Z}_{\langle 5 \rangle}$, but if $\frac{5}{1}$ were to be the zero element then we'd have $\frac{5}{1}=\frac{0}{1}$ and so $5u=0$ for some $u\in\mathbb{Z}\setminus 5\mathbb{Z}$. But this is an equation in $\mathbb{Z}$, and as $u\in\mathbb{Z}\setminus 5\mathbb{Z}$ we have $u\neq 0$, so since $\mathbb{Z}$ is a domain this is impossible. Thus $\frac{5}{1}\in5\mathbb{Z}_{\langle 5 \rangle}$ is non-zero and so $5\mathbb{Z}_{(5)}$ is not the zero ideal. Please correct me if I have some gap in understanding! – Martin May 5 at 19:29
• You said yourself it's a local ring. It's actually a field. As such it has the zero ideal as its unique maximal ideal. Furthermore it has characteristic $5$. So $5r=0$ for any $r$. – Chris Custer May 5 at 21:56