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My question actually follows from this one:

Chinese Remainder Theorem for Rings

What I don't understand is why is it necessary for $I+J=R$, in order for

$$ \varphi\colon R\to R/I\times R/J:a\mapsto (a+I,a+J) $$

to be surjective. This map seems surjective by construction (I worked through the same question myself and came up with the same homomorphism. But I didn't seem to require that $I+J=R$ in my answer).

Quoting Ben West's answer, he says:

For any $r,s\in R$, we can write $r=r_i+r_j$ and $s=s_i+s_j$, where $r_i,s_i\in I$, etc. Let $a=r_i+s_j$, so $a-r=s_j-r_j\in J$, and $a-s=r_i-s_i\in I$. In particular, $a+I=s+I$, and $a+J=r+J$. This shows that the map $$ \varphi\colon R\to R/I\times R/J:a\mapsto (a+I,a+J) $$ is surjective.

Why does $a-s \in I $ $\Rightarrow a+I = s+I $ in the above argument?

Thanks so much in advance!

Community
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JJJ
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2 Answers2

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If $a - s\in I$, then $a - s = x$ for some $x\in I$. Then $a = s + x$. Thus, for every $y \in I$, $a + y = s + (x + y) \in s + I$, showing that $a + I \subseteq s + I$. Similarly, $s + I \subseteq a + I$. Hence $a + I = s + I$.

If $I + J \neq R$, $\phi$ is not surjective. Just to illustrate, if $R = \Bbb Z$, $I = (3)$, and $J = (6)$, then $(2 + I, 3 + J)$ is not in the image of $\phi$. Otherwise, $\phi(a) = (2 + I, 3 + J)$ for some $a\in \Bbb Z$, which implies $a \equiv 2\pmod{3}$ and $a \equiv 3 \pmod{6}$. Since $a \equiv 3\pmod{6}$, $a = 3 + 6j$ for some $j\in \Bbb Z$. Since $3 + 6j \equiv 0 \pmod{3}$, then $a \equiv 0\pmod{3}$. This contradicts the congruence $a \equiv 2\pmod{3}$.

kobe
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  • Thanks, I think I should have went one step further. I now see that $a+I = s+I$ and $a+J=r+J$, how does this then show that the map is surjective? Is it because then for any two arbitrary elements in $R$, there exists an element $a \in R$ which maps to that particular pair? – JJJ Apr 11 '15 at 12:44
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    So given $r,s \in R$, $\phi(a) = (r + I, s + J)$, where $a = r_i + s_j$ in Ben's construction. This proves $\phi$ is surjective. Makes sense? – kobe Apr 11 '15 at 12:49
  • Ahh yes that makes sense. Thanks! – JJJ Apr 11 '15 at 13:11
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The map $\varphi\colon R\colon R/I\times R/J$ defined by $r\mapsto(r+I,r+J)$ is a ring homomorphism with kernel $I\cap J$, so it induces an injective homomorphism $$ R/(I\cap J)\to R/I\times R/J $$ which is also surjective if and only if $\varphi$ is surjective.

So, assume $\varphi$ is surjective. This means that $(1+I,0+J)=\varphi(r)$ for some $r$, that is $$ r+I=1+I,\quad r+J=0+J $$ Thus $1-r=x\in I$ and $r=y\in J$, which means $1=x+y\in I+J$, forcing $I+J=R$.

The converse, that is, $I+J=R$ implies $\varphi$ surjective, is easy.

egreg
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  • Just wondering, is this an alternative answer to this question which you have answered? Or are they different? http://math.stackexchange.com/questions/1152991/converse-to-chinese-remainder-theorem/1244165#1244165 – user10024395 Apr 21 '15 at 01:55
  • @Aha They are different, because here the canonical homomorphism is used. The other is more general. – egreg Apr 21 '15 at 10:27