Vrftsjtryk

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,trbrace $; is
: $ mathcal{J}(mathbb{Z}_m)=bigcap_{ mathfrak{m}vartriangleleftcdot mathbb{Z}_m}mathfrak{m}=bigcap_{iin I}dfrac{p_imathbb{Z}}{mmathbb{Z}}=bigcap_{iin I}([p_i]_{mmathbb{Z}})=bigl(text{lcd} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})bigl)=left(dfrac{[p_1]_{mmathbb{Z}}cdots[p_t]_{mmathbb{Z}}}{text{GCD} ([p_1]_{mmathbb{Z}},ldots,[p_t]_{mmathbb{Z}})}right)= ([p_1cdots p_t]_{mmathbb{Z}})=dfrac{p_1cdots p_tmathbb{Z}}{mmathbb{Z}}=dfrac{sqrt{mmathbb{Z}}}{mmathbb{Z}}=mathcal{N}(mathbb{Z}/mmathbb{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 $P=nmathbb{Z}/mmathbb{Z}$ for some $ninmathbb{Z}$ by the correspondence theorem (3rd isomorphism theorem) for rings. But note that $nmathbb{Z}$ is a prime ideal in $mathbb{Z}$ since $mathbb{Z}_{m}/P$ is an integral domain and
$$mathbb{Z}_{m}/P=(mathbb{Z}/mmathbb{Z})/(nmathbb{Z}/mmathbb{Z})simeqmathbb{Z}/nmathbb{Z}$$
by the correspondence theorem. But this implies that $mathbb{Z}/nmathbb{Z}$ 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
$$mathcal{J}(mathbb{Z}_{m})=bigcaplimits_{mathfrak{m}text{ is maximal}}mathfrak{m}=bigcaplimits_{mathfrak{p}text{ is prime}}mathfrak{p}=mathcal{N}(mathbb{Z}_{m}).$$