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If $R$ contains two ideals $B$ and $C$ with $B+C=R$ and $B \cap C=0$, then $B$ and $C$ are rings and $R\cong B\times C$

I tried to prove it using definition of subrings, that is, B and C is closed under identity, subtraction and multiplication. But then I encountered a problem: If they are rings, then they both contain the multiplicative unit 1, but it contradicts with the fact that B intersect C is 0.

If they have different multiplicative units, then they can't be subrings of R since the unit is not preserved. I'm stuck on this....

What do you guys suggest? Any help will be appreciated!

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Do your rings have to have multiplicative identity? Note that a multiplicative identity for $B$ doesn't have to be the multiplicative identity of the whole ring. – Arturo Magidin Nov 17 '11 at 5:35
So, B and C have to be rings in their own right. They do not necessarily have to be subrings of R. For example, if $e$ is a non-trivial idempotent of a ring $A$ (that is $e$ is an idempotent not equal to 0,1), then it is true that the ideals $Ae$ and $A(1-e)$ are both rings in their own right, but not necessarily subrings of $A$. In fact in this case you do have $A \cong Ae \times A(1-e)$. – Rankeya Nov 17 '11 at 5:37
up vote 4 down vote accepted

Do your rings have to have a multiplicative identity? If not, then every ideal is automatically a subring. If yes, then notice that a multiplicative identity for $B$ doesn't have to be the multiplicative identity of $R$: cheating and taking what "the answer" should be, consider $\mathbb{Z}\times\mathbb{Z}$, $B=\mathbb{Z}\times\{0\}$, $C=\{0\}\times\mathbb{Z}$. Note that $(1,0)$ is an identity for $B$, but is not the identity of $R$. Now, using that example as a mental guide:

Since $B+C=R$, there exists $f\in B$ and $e\in C$ such that $f+e=1$.

Show that $bf = fb = b$ for all $b\in B$ and $ec=ce = c$ for all $c\in C$. That will show that each of $B$ and $C$ have a multiplicative identity, and hence are rings (all other requirements for being a ring are met by virtue of them being ideals).

Now, map $B\times C$ to $R$ by $(b,c)\longmapsto b+c$. Show this is an onto ring-homomorphisms, and is one-to-one.

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thanks!I got your idea, but I wonder if it's supposed to be be=eb=b because in the Z example, the identity for B is (1,0) – Scharfschütze Nov 17 '11 at 7:40
@Scharfschütze: Ehr, yes, I got my $e$ and $f$'s reversed in the previous line. I'll fix it. – Arturo Magidin Nov 17 '11 at 14:03

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