$\mathbb{Z}\mathrm{/}12\mathbb{Z}$ splits as the product of its two local Artinian factors

The ideals $(3)$ and $(4)$ of $\mathbb{Z}$ are comaximal, so Chinese remainder theorem for pairwise comaximal ideals gives $\mathbb{Z}\mathrm{/}12\mathbb{Z}\cong \mathbb{Z}\mathrm{/}3\mathbb{Z}\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. The first factor is a field, hence local, and the second has unique proper nonzero ideal $(2)$, so it is also local.

Let $P$ be a prime ideal of $\mathbb{Z}\mathrm{/}3\mathbb{Z}\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. Since $(1,0)(0,1)=(0,0)\in P$, primality forces $(1,0)\in P$ or $(0,1)\in P$. In the first case every $(a,0)=(a,0)(1,0)$ lies in $P$, so $P=\mathbb{Z}\mathrm{/}3\mathbb{Z}\times Q$ for an ideal $Q$ of $\mathbb{Z}\mathrm{/}4\mathbb{Z}$; primality then forces $Q=(2)$. In the second case $P=0\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$. Under the inverse of the isomorphism in step 1.1, these two primes pull back to $(2)\mathrm{/}(12)$ and $(3)\mathrm{/}(12)$ in $\mathbb{Z}\mathrm{/}12\mathbb{Z}$.

So $\mathbb{Z}\mathrm{/}12\mathbb{Z}$ is exhibited concretely as the product of its two local Artinian factors $\mathbb{Z}\mathrm{/}3\mathbb{Z}$ and $\mathbb{Z}\mathrm{/}4\mathbb{Z}$, and its two prime ideals are exactly the two coordinate pullbacks found in step 2.1.