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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.

Could anyone advise on my answer? Thank you.

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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.

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  • $\begingroup$ $\ \ \mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5} \cong \mathbb{Z}_{7}.$ Since $\mathbb{Z}_{7}$ is field, so is $\mathbb{Z}_{7^5} / 7\mathbb{Z}_{7^5}$. Hence, $(7)$ is maximal ideal. $\endgroup$ – Alexy Vincenzo May 2 '14 at 6:42
  • $\begingroup$ "Highly nonstandard" is probably relative to one's background. The notation is exceedingly common (or else why would it constantly appear here) and totally understandable in context, most of the time. $\endgroup$ – rschwieb Oct 9 '14 at 11:39

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