I want to proove (finding explicit formulas) that the Jacobson radical of $\mathbb{Z}_m$ equals its nilradical: i.e. $ \mathcal{N}(\mathbb{Z}_m)=\mathcal{J}(\mathbb{Z}_m) $. I prooved a formula (following Atiyah - Mac Donald, Introduction to commutative algebra - pag. 6 - Examples 1) for the ring $\mathbb{Z}_m$, where $m>1$, $ m=p_1^{s_1}\cdots p_t^{s_t} $ and $ I=\lbrace 1,\ldots,t\rbrace $; is : $ \mathcal{J}(\mathbb{Z}_m)=\bigcap_{ \mathfrak{m}\vartriangleleft\cdot \mathbb{Z}_m}\mathfrak{m}=\bigcap_{i\in I}\dfrac{p_i\mathbb{Z}}{m\mathbb{Z}}=\bigcap_{i\in I}([p_i]_{m\mathbb{Z}})=\bigl(\text{lcd} ([p_1]_{m\mathbb{Z}},\ldots,[p_t]_{m\mathbb{Z}})\bigl)=\left(\dfrac{[p_1]_{m\mathbb{Z}}\cdots[p_t]_{m\mathbb{Z}}}{\text{GCD} ([p_1]_{m\mathbb{Z}},\ldots,[p_t]_{m\mathbb{Z}})}\right)= ([p_1\cdots p_t]_{m\mathbb{Z}})=\dfrac{p_1\cdots p_t\mathbb{Z}}{m\mathbb{Z}}=\dfrac{\sqrt{m\mathbb{Z}}}{m\mathbb{Z}}=\mathcal{N}(\mathbb{Z}/m\mathbb{Z})=\mathcal{N}(\mathbb{Z}_m) $
Is it correct?
Your formula is correct and in fact we have that $\mathcal{J}(\mathbb{Z}_{m})=\mathcal{N}(\mathbb{Z}_{m})$. Indeed, if $P$ is a non-zero prime ideal in $\mathbb{Z}_{m}$ then $P=n\mathbb{Z}/m\mathbb{Z}$ for some $n\in\mathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $n\mathbb{Z}$ is a prime ideal in $\mathbb{Z}$ since $\mathbb{Z}_{m}/P$ is an integral domain and $$\mathbb{Z}_{m}/P=(\mathbb{Z}/m\mathbb{Z})/(n\mathbb{Z}/m\mathbb{Z})\simeq\mathbb{Z}/n\mathbb{Z}$$ by the correspondence theorem. But this implies that $\mathbb{Z}/n\mathbb{Z}$ is a field because every non-zero prime ideal in $\mathbb{Z}$ is maximal and therefore $P$ is a maximal ideal in $\mathbb{Z}_{m}$ as well. Thus $$\mathcal{J}(\mathbb{Z}_{m})=\bigcap\limits_{\mathfrak{m}\text{ is maximal}}\mathfrak{m}=\bigcap\limits_{\mathfrak{p}\text{ is prime}}\mathfrak{p}=\mathcal{N}(\mathbb{Z}_{m}).$$