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Let $A_1, \dots , A_n$ be commutative unitary Rings and $A = \prod_{i=1}^{n} A_i$. Then every prime Ideal $\frak{p}$ $\subset A$ is of the form $\pi_i^{-1}(\frak{p}_i)$ where $\pi_i: A \to A_i$ are the canonical projections and $\frak{p_i}$ $\subset A_i$ is prime.

I know that every Ideal in $A$ is a direct product of Ideals in the $A_i$, so we have $\frak{p}$ $= \frak{a}_i \times \dots \times \frak{a}_n$ for some Ideals $\frak{a}_i$ $\subset A_i$. They are also prime since $\frak{p}$ is prime. Now, since the $\pi_i$ are surjective $\pi_i(\frak{p})$ $\subset A_i$ is prime as well. Then we have $\pi_i^{-1}(\pi_i(\frak{p})) \supset \frak{p}$. Now I don't know how to finish the proof.

Hints and/or improvements are greatly appreciated!

Shuster
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    Your sentence "They are also prime since $\mathfrak{p}$ is prime" is not correct. Also notice that this contradicts the claim which you want to prove since $$\pi_i^{-1}(\mathfrak{p}i) = A_1 \times \cdots \times A{i-1} \times \mathfrak{p}i \times A{i+1} \times \cdots \times A_n.$$ – Martin Brandenburg Apr 17 '21 at 17:02
  • Please feel free to correct me: Let $xy \in \frak{p}$, that means $x_iy_i \in \frak{a}_i$ for every $i$. Since $\frak{p}$ is prime, we have $x \in \frak{p}$ or $y \in \frak{p}$, that means $x_i \in \frak{a}_i$ or $y_i \in \frak{a}_i$, so the $\frak{a}_i$ should be prime. – Shuster Apr 17 '21 at 17:05
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    Prime ideals are assumed to be proper, so from the above description of the prime $\mathfrak{p}$, you see that the claim is not correct. Also, your proof starts with the wrong assumption. When you want to prove $\mathfrak{a}_i$ is prime, start with $xy \in \mathfrak{a}_i$. – Martin Brandenburg Apr 17 '21 at 17:08
  • You're right, I'm sorry. Totally didn't notice that. Thanks for clarification! – Shuster Apr 17 '21 at 17:09
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    For example, $0 \times \mathbb{Z}$ is a prime ideal of $\mathbb{Z} \times \mathbb{Z}$ (since the quotient is $\cong \mathbb{Z}$). But $\mathbb{Z}$ is not a prime ideal of $\mathbb{Z}$. – Martin Brandenburg Apr 17 '21 at 17:10
  • While historically $\subset$ has been used to mean $\subseteq,$ that was probably more a case of typesetting costs. It is unclear in $\mathfrak a_i\subset A_i$ whether you mean $\subseteq$ or $\subsetneq.$ – Thomas Andrews Apr 17 '21 at 17:37

2 Answers2

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You can derive the proof from the following facts:

  1. An ideal $\mathfrak{a} \subseteq A$ is prime iff $A/\mathfrak{a}$ is an integral domain.

  2. There is an isomorphism of rings $(\prod_i A_i) / (\prod_i \mathfrak{a}_i) \cong \prod_i A_i / \mathfrak{a_i}$

  1. A product $\prod_i A_i$ is an integral domain iff $A_i=0$ for all indices $i$ except for exactly one index $j$, and for this index $A_j$ is an integral domain.
Martin Brandenburg
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Assume $\mathfrak a_i\neq A_i,\mathfrak a_j\neq A_j$ for $i\neq j.$ Then $1_i\in A_i\setminus \mathfrak a_i$ and $1_j\in A_j\setminus \mathfrak a_j.$

Define $a=(a_k)_{k=1}^n,b=(b_k)_{k=1}^n\in A$ as:

$$a_k=\begin{cases}0&k\neq i\\ 1_i&k=i \end{cases}$$

$$b_k=\begin{cases}0&k\neq j\\ 1_j&k=j \end{cases}$$

Then $ab=0\in\mathfrak p,$ but neither $a$ nor $b$ is in $\mathfrak p,$ so $\mathfrak p$ is not prime.

So if $\mathfrak p=\prod \mathfrak a_i$ is a prime ideal, then $\mathfrak a_i=A_i$ must be true for all but one $i.$

Thomas Andrews
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