Prime Ideals of Products of $\mathbb Z_p$

Question

I know that the Prime Ideals of $\mathbb Z_p\times\mathbb Z_p$ are $\langle(1,0)\rangle $ and $\langle(0,1)\rangle$, but what would it be for $\mathbb Z_p\times\mathbb Z_p\times\mathbb Z_p$? Would it be like $\langle0,1,0\rangle$ or $\langle1,0,1\rangle $ or something else? I believe it should be the second option, but I'm not sure why. I also assume the pattern would continue for more products.

Answer 1

The prime ideals in a finite product of rings $R=\prod \limits_{i=1}^n R_i$ are all of the form $\mathfrak{p}_k \times \prod \limits_{i=1,i\neq k}^n R_i$, where $\mathfrak{p}_k \subseteq R_k$ is a prime ideal. That these are prime ideals follows directly from the fact that they are the kernels of surjective ring-homomorphisms of the form $$\prod \limits_{i=1}^n R_i \rightarrow R_k/\mathfrak{p}_k, (r_1,...,r_n) \mapsto r_k + \mathfrak{p}_k.$$ Conversely, if $\mathfrak{p}\subseteq R =\prod \limits_{i=1}^n R_i$ is a prime ideal, its images $\pi_i(\mathfrak{p})\subseteq R_i$ under the canonical projections give rise to a homomorphism $$ \phi:\prod\limits_{i=1}^n R_i \rightarrow \prod\limits_{i=1}^nR_i/\pi_i(\mathfrak{p}),(r_1,...,r_n)\mapsto (r_1+\pi_1(\mathfrak{p}),...,r_n+\pi_n(\mathfrak{p}))$$ which sends every element of $\mathfrak{p}$ to zero and thus satisfies $\mathfrak{p} \subseteq \ker \phi = \prod\limits_{i=1}^n\pi_i(\mathfrak{p})$. But on the other hand we also have $\prod\limits_{i=1}^n\pi_i(\mathfrak{p})\subseteq \mathfrak{p}$, since for every element $p_k \in \pi_k(\mathfrak{p})$ there is an element $(p_1,...,p_n)\in \mathfrak{p}$ whose $k$-th component is $p_k$ and by multiplying with $(0,..,1_k,...,0)$ we have that $(0,...,p_k,...0)\in \mathfrak{p}$. Any general element of $\prod\limits_{i=1}^n\pi_i(\mathfrak{p})$ is just a sum of those.

Finally suppose there is more than one index, where $\pi_i(\mathfrak{p}) \neq R_i$. Then the quotient $R/\mathfrak{p} = \prod\limits_{i=1}^n R_i/\pi_i(\mathfrak{p})$ is a product of nontrivial rings, hence not an integral domain. This would contradict primality of $\mathfrak{p}\subseteq R$.

Note: primality of $\mathfrak{p}$ only enters in the proof by contradiction in the end. We have in fact proven that any ideal in a finite product of rings is just a product of ideals.