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Consider $\mathbf Z/6\mathbf Z$ as a ring.

It is not an integral domain since it contains zero-divisors, such as the element $[3]$ for example. Note that $[3]$ is not irreducible ($[3]^2 = [3]$), yet $\langle[3]\rangle$ as an ideal is maximal, since modding out by $\langle[3]\rangle$ produces the field $\mathbf F_2$. Since $\mathbf F_2$ is a field, it is an integral domain, hence $\langle[3]\rangle$ is a prime ideal.

Does this imply that $[3]$ is a prime element of $\mathbf Z/6\mathbf Z$?

Further, does it even make sense to talk about prime elements of non-domains?

node196884
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  • Factorization theory is more complicated in non-domains, i.e. in rings with nontrivial zero-divisors. Basic notions such as associate and irreducible bifurcate into a few inequivalent notions - see the literature cited in the linked dupe. But the notion of prime elements $p$ remains the same, i.e. $(p)$ is a prime ideal. – Bill Dubuque Jul 02 '24 at 19:41
  • quite helpful, thanks – node196884 Jul 02 '24 at 20:37

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