# An attempt to verify if $\mathbb{Z}_{7^5}$ a local ring with unique maximal ideal $(7) \ .$

A commutative ring $R$ with identity is called a local ring if there exists unique maximal ideal in $R.$ Hence, is $\mathbb{Z}_{7^5}$ a local ring with unique maximal ideal $(7) \ ?$

Here is my answer: No. $\ \ \mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5} \cong \mathbb{Z}_{7^4}.$ $\ \ \mathbb{Z}_{7^4}$ is not a field, so $\mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5}$ is also not a field. Hence, $(7) = 7\mathbb{Z}_{7^5}$ is not maximal ideal.

You're using highly nonstandard notation, but I assume by $\mathbb{Z}_{7^5}$ you mean $\mathbb{Z}/7^5\mathbb{Z}$. Usually you reserve subscripts of $\mathbb{Z}$ to denote the $p$-adic integers.
Anyway, you need to think about what $7\mathbb{Z}/7^5\mathbb{Z}$ really is. It might help to do a simpler example, say...
is $\mathbb{Z}/16\mathbb{Z}$ a local ring? What are the elements of this ring? (note there are 16) What are the elements of the ideal $2(\mathbb{Z}/16\mathbb{Z})$? (note there are 8). How many elements are in the quotient? (hint: not very many).
Spoiler: $\mathbb{Z}/p^n\mathbb{Z}$ is a (nonreduced) local ring for any prime $p$ and $n\ge 1$, with maximal ideal $p$. In fact, not only does it have a unique maximal ideal, its unique maximal ideal is also its unique prime ideal.
• $\ \ \mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5} \cong \mathbb{Z}_{7}.$ Since $\mathbb{Z}_{7}$ is field, so is $\mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5}$. Hence, $(7)$ is maximal ideal. – Alexy Vincenzo May 2 '14 at 6:42