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Let $R_1, \ldots, R_n$ be rings and consider the cartesian product $R_1\times\ldots\times R_n$ as a ring under componentwise addition and multiplication. Show there is a canonical bijection $$\operatorname{Spec}(R_1\times\ldots\times R_n)\xrightarrow{\ \sim\ }\coprod_{i=1}^n\operatorname{Spec}R_i$$ and a similar one for spectra of maximal ideals.

Attempt:

Let $a=(a_1,\ldots,a_n)\in \operatorname{Spec}(R_1\times\cdots\times\ R_n)$, then $(R_1\times\cdots\times R_n)/a=(R_1/a_1,\ldots,R_n/a_n)$ is an integral domain. Suppose there exists nonzero $b_i,c_i\in R_i/a_i$ for all $1\leq i\leq n$ such that $b_ic_i=0$, then $b,c\in R/a$ are nonzero elements such that $bc=0$, so at least one of the factors is an integral domain.

This is what I can get. I don't know how to find the bijection. I think the L.H.S. is bigger than the R.H.S.

Edit: This is from a more elementary book of commutative algebra, so I didn't know the map in the Atiyah-Macdonald book.

Dasheng Wang
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    What do you mean by $a=(a_1,\ldots, a_n)$? Are the $a_i$ elements of the rings $R_i$? Are they ideals? – Sebastian Monnet Feb 22 '22 at 18:02
  • @SebastianMonnet Yes, I forgot to add that ideals of $R_1\times\cdots\times\ R_n$ have the form of $a_1\times\cdots \times a_n$ where $a_i$ are ideals of $R_i$, this is proven in the previous problem. – Dasheng Wang Feb 22 '22 at 18:05
  • You can show that prime ideals in $R_1\times\cdots\times R_n$ are of the form $R_1\times\cdots \times p_i\times\cdots\times R_n$ for some prime ideal $p_i\subseteq R_i$ (so just one of the components is a prime, and the other components are the unit ideal). – Dave Feb 22 '22 at 18:11
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    Duplicate of Disjoint Union of Spectra – KReiser Feb 22 '22 at 20:57

1 Answers1

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Write $R = R_1\times\cdot \times R_n$. Let $P = P_1\times \ldots \times P_n$ be a prime ideal of $R$.

Claim: The ideals $P_i$ are either prime or equal to $R_i$.

Proof: Without loss of generality take $i=1$. Let $x,y \in R_1$ with $xy \in P_1$. Then $(x,0,\ldots, 0)\cdot (y,0,\ldots, 0) \in P$ one of $(x,0,\ldots, 0)$ and $(y,0,\ldots, 0)$ is in $P$, hence one of $x$ and $y$ is in $P_1$. Therefore $P_1$ is prime or $P_1 = R_1$.

Claim: At most one of the $P_i$ is not equal to $R_i$.

Proof: Suppose not. Without loss of generality suppose that $P_1\neq R_1$ and $P_2 \neq R_2$. Then $(1,0,\ldots,0), (0,1,0,\ldots, 0) \in R\setminus P$, but their product is in $P$, so $P$ cannot be prime.

Therefore, a prime ideal of $R$ is basically the same thing as a prime ideal of one of the rings $R_i$, which is an element of the right-hand-side in your question.

Sebastian Monnet
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    Not all the $P_i$ are prime (a condition of being prime is that the ideals are proper). And I think your proof that the $P_i$ are prime is backwards: we want $ab\in P_1\implies a\in P_1\lor b\in P_1$, not $x\in P_1,r\in R_1\implies xr\in P_1$ (the latter is the condition of just being an ideal). – Dave Feb 22 '22 at 18:43
  • $\mathbb Z\times 2\mathbb Z$ is a prime ideal of $\mathbb Z\times \mathbb Z$, but $\mathbb Z$ is not a prime ideal of $\mathbb Z$. – Dasheng Wang Feb 22 '22 at 18:55
  • Yes, @Dave is absolutely right. I wrote my answer in a hurry and totally butchered that part of it. Have now fixed it. – Sebastian Monnet Feb 22 '22 at 18:55