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Let $R$ and $S$ be two rings, then $K$ is an ideal in $R\times S$ if and only if there are $K_1$ and $K_2$ such that $K_1$ is an ideal of $R$ and $K_2$ is an ideal of $S$ such that $K = K_1 \times K_2$.

How do you show that $K_1 = \{a \in R \mid (a,0) \in K\}$ is an ideal of $R$ and $K_2 = \{b \in S \mid (0,b) \in K\}$ is an ideal of S, and the converse too.

Thank you very much.

jgon
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Newbie Math
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  • Take $K_1$ and $K_2$ as the projections of $K$ and check that they are ideals and $K=K_1\times K_2$. You can easily check that $K\subset K_1\times K_2$. Showing the reverse inclusion requires some properties of ideals. – Hanul Jeon May 14 '15 at 01:12
  • how to check K1 and also K2 are ideals in form as (a,0) ∈ K, (0,b) ∈ K ? – Newbie Math May 14 '15 at 01:21

1 Answers1

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For the inclusion $K_1 \times K_2 \subset K$, let $(a,b) \in K_1 \times K_2$, then $(a,0) \in K$ and $(0,b) \in K$.

For the reverse inclusion, let $(a,b) \in K$, then $(1,0) \cdot (a,b)=(a,0) \in K$ and $(0,1) \cdot (a,b)=(0,b) \in K$. Note that I assume $1 \in R,S$.

ET93
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  • how to check K1 and also K2 are ideals in form as (a,0) ∈ K, (0,b) ∈ K ? – Newbie Math May 14 '15 at 01:23
  • Take two elements in $a,b \in K_1$ What is their sum and product in K_1? What about their sum and product in $K$ as $(a,0),(b,0)$. Take $r \in R$. Is $(ra,0) \in K$? Notice $(r,0) \cdot (a,0)=(ra,0)$. – ET93 May 14 '15 at 01:28
  • Ok, thank very much. I've got it – Newbie Math May 14 '15 at 01:40