# Rings with decomposition are sums of ideals generated by idempotents [duplicate]

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I'm working on a question out of T.Y. Lam's book that has me thrown.

Let $$B_1 \dots, B_n$$ be left ideals (resp. ideals) in a ring $$R$$. Show that $$R=B_1 \oplus \dots \oplus B_n$$ iff there exists idempotents (resp. central idempotent) $$e_1, \dots ,e_n$$ with sum 1 such that $$e_ie_j=0$$ whenever $$i \neq j$$, and $$B_i=R e_i$$ for all $$i$$. In the case where the $$B_i$$'s are ideals, if $$R=B_1 \oplus \dots \oplus B_n$$, then each $$B_i$$ is a ring with identity $$e_i$$, and we have an isomorphism between $$R$$ and the direct product of rings $$B_1 \times \dots \times B_n$$. Show that any isomorphism of $$R$$ with a finite direct product of rings arises in this was.

# Attempt

Supposing we have such idempotents there is an obvious map $$R \to \oplus Re_i$$ since $$r=r \cdot 1=r \sum e_i$$. Hence, $$r \mapsto r \sum e_i$$. This map is easily shown to be bijective, however, demonstrating that this is a homomorphism seems to be problematic. Consider

$$r_1r_2 \mapsto r_1r_2(\sum e_i) \text{ but } r_1(\sum e_i)r_2(\sum e_i) \neq r_1r_2 \sum e_i???$$

so perhaps this isn't the correct map?

Conversely, supposing we have such a decomposition then clearly $$1=\sum b_i$$ so let $$e_i = b_i$$. I'm guessing the fact that 1 is trivially an idempotent will show that the $$e_i$$ are but I'm stuck there. We have

$$1=1^2=(\sum e_i)^2$$

and similar to the above the map $$r=r*1=r\sum e_i$$ jumps out at me.

## marked as duplicate by Aaron Zolotor, Lord Shark the Unknown abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 18 '18 at 4:28

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## 1 Answer

Let's try the left ideal case.

Suppose $$R=B_1\oplus B_2\oplus\dots\oplus B_n$$. Then $$1=\sum_{j=1}^n e_j$$. Moreover, for $$1\le i\le n$$, $$e_i=e_i1=\sum_{j=1}^n e_ie_j$$ Since $$e_ie_j\in B_j$$, we get $$e_ie_j=\begin{cases} e_i & i=j \\[4px] 0 & i\ne j \end{cases}$$ If $$x\in B_i$$, then $$x=x1=\sum_{j=1}^n xe_j$$ so we get $$xe_j=\begin{cases} x & i=j \\[4px] 0 & i\ne j \end{cases}$$ In particular, $$B_j=B_je_j=Re_j$$.

Suppose conversely that we have idempotents $$e_1,\dots,e_n$$ with $$e_1+\dots+e_n=1$$ and $$e_ie_j=0$$ for $$i\ne j$$.

The map $$f\colon R\to Re_1\oplus Re_2\oplus\dots\oplus Re_n$$ defined by $$f(r)=\sum_{j=1}^n re_j$$ is clearly a homomorphism of $$R$$-modules. If $$r\in\ker f$$, then $$r1=0$$. What about surjectivity? Suppose $$r_j\in Re_j$$, for $$j=1,2,\dots,n$$. Then $$r_je_j=r_j$$ and if we set $$r=r_1+r_2+\dots+r_n$$, we have $$f(r)=\sum_{j=1}^n re_j= \sum_{j=1}^n\biggl(\,\sum_{i=1}^n r_i\biggr)e_j= \sum_{j=1}^n\biggl(\,\sum_{i=1}^n r_ie_ie_j\biggr)= \sum_{j=1}^n r_je_j=\sum_{j=1}^n r_j$$

The two-sided ideal case is similar. The map $$f$$, in this case, is also a ring homomorphism, with each $$B_i$$ being a ring with unit $$e_i$$.

• Is an R-module $isomorphism$ strong enough to show these two objects are isomorphic as rings? – Aaron Zolotor Oct 16 '18 at 1:45
• @RhythmInk No, it isn't. – egreg Oct 16 '18 at 7:48
• That sum doesn't indicate necessarily $e_ie_j=0$ for $i \neq j$. I know that will require the fact that the $B_i$ are all pairwise trivially intersected but I don't see it at the moment. – Aaron Zolotor Oct 18 '18 at 1:02