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First of all, when I say that $(R, +, \cdot)$ is a ring, I mean $R$ commutative ring with identity and I am using the definition that $(S, +, \cdot)$ is a subring of $R$ if $(S, +)$ is a subgroup of $(R, +)$ and S is closed under $\cdot$ and has an identity but that identity need not be the same as the idenity of $R$.

I was able to show that of if $\mathbb{Z}_{d}$ is a subring then $d\mid m$ and I expected the converse to be true as well (if $d \mid m$ then $\mathbb{Z}_{d}$ is a subring of $\mathbb{Z}_{m}$) but that fails if we take $m = 12$ and $d = 6$.

Is there a way to determine if $\mathbb{Z}_{d}$ is a subring of $\mathbb{Z}_{m}$ by just looking at the values of $d$ and $m$?

Ethan Kharitonov
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    Hint: Let $p$ be a prime, can you show that the pair $m=p^2$ and $d=p$ fails? Why does this pair fail? What happens when $d$ and $m/d$ are relatively prime? – Michael Burr Aug 19 '23 at 20:27
  • At the very least, you will need the image $x$ of the multiplicative identity of the subring to satisfy $x^2=x$ in $\mathbb{Z}/m$. Take a look at https://math.stackexchange.com/questions/45747/how-many-idempotent-elements-does-the-ring-bf-z-n-contain and https://math.stackexchange.com/questions/1661650/number-of-solutions-for-x2-equiv-x-pmod-m. See also https://math.stackexchange.com/questions/1575074/find-all-positive-integer-n-such-that-mathbbz-n-contains-a-subring-isom?rq=1. – John Palmieri Aug 19 '23 at 20:30

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