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I am trying to study the proof of the Chinese remainder theorem from the book Dummit and Foote (Section 7.6, page 266, third edition). The theorem itself is to show that $R/(A_1A_2\ldots A_k) \cong R/A_1 \times \ldots \times R/A_k) $ if the ideals $A_1, \ldots,A_k$ are comaximal, i.e $A_i +A_j=R$ if $i \neq j$. The proof is by induction and I understand the case $k=2$. Now they are trying to show that $A_1$ and $(A_2 A_3 \ldots A_k)$ are comaximal and I don't fully understand this argument. It is claimed that for $i \in \{ 2,3,\ldots,k\}$ there are $x_i \in A_1$ and $y_i \in A_i$ such that $x_i + y_i =1$ (I guess this follows from the hypothesis that the ideals are comaximal, and that $x_i +y_i \equiv y_i (\mbox{mod} A_1)$ (which is because $x_i + y_i + A_1 = y_i + A_1$ since $x_i \in A_1$ ). Now they say $1= (x_2+y_2) \ldots (x_k+y_k)$ is an element in $A_1 + (A_2 \ldots A_k)$.

I would be grateful if somebody can kindly elaborate on this last statement. I also don't see where they use the fact that $x_i +y_i \equiv y_i (\mbox{mod} A_1)$

Thank you.

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I really like Jiangwei's hint. You can also think about the terms you get when you expand that product: every term besides $y_2 \cdots y_k$ is contained in $A_1$. –  Dylan Moreland Jul 15 '11 at 15:01
That seems a bit easier. But in that case do we actually need $x_i + y_i \cong y_i (mod A_i)$. –  Shibi Vasudevan Jul 15 '11 at 15:27
No, but there are often many ways to prove these simple lemmas. I think Bill's answer will clarify the book's intent. –  Dylan Moreland Jul 15 '11 at 17:22
Thanks for your answer. –  Shibi Vasudevan Jul 15 '11 at 19:56
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2 Answers

up vote 3 down vote accepted

Say $I$ and $J$ are two ideals in a ring $R$, and $a \equiv b \mod I$, where $b\in J$, then does it follow that $a\in I+J$?

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Thanks for the answer. $a \cong b \mbox{mod} I$ means that $a-b \in I$ this means that $a=a-b + b \in I + J$. But now, how do I use this? Do I put $ a= x_i + y_i $ and $b= y_i$ I get that $x_i \in A_1 + A_i$. How does the statement then follow? –  Shibi Vasudevan Jul 15 '11 at 14:49
@ Jiangwei: Could you please elaborate the hint a little bit more. I don't fully understand it. Thanks. –  Shibi Vasudevan Jul 15 '11 at 15:28
@Chandru: Why? That's Shibi's call and Shibi's alone. There are good reasons to wait a bit longer than only an hour. Also, Shibi didn't get it yet, so why on earth accept? –  t.b. Jul 15 '11 at 16:36
@Theo: Oh, I just read the first sentence of his comment:"Thanks for the answer." and wrote up this comment, which now I delete because, it was my mistake of not reading the whole comment. –  user9413 Jul 15 '11 at 16:57
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HINT $\rm\ \ mod\ A_1:\ \ x_i\equiv 0 \ \Rightarrow\ (x_2+y_2)\:\cdots\:(x_k+y_k)\ \equiv\ y_2\:\cdots\:y_k \in A_2\:\cdots\:A_k$

Alternatively use $\rm\ (A + B)\ (A + C)\ \subseteq\ A + B\ C\ $ and induction.

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Thanks a lot for the hint –  Shibi Vasudevan Jul 15 '11 at 19:57
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