Cfrgtkky

Let $R$ be a ring. The Chinese Remainder Theorem asserts that if $J_1, ldots, J_nlhd R$ are bilateral coprime ideals ( i.e. $J_i+J_k=R, forall ineq k$) then the map $$varphi:ain Rlongmapsto (a+J_1,...,a+J_n)in R/J_1oplus ...oplus R/J_n $$ is a surjective ring morphism, such that $$ker (varphi)=displaystyle bigcap _{1leq ileq n} J_i.$$ Thus we have a ring isomorphism (and thus a R-module isomorphism) $$frac{R}{displaystyle bigcap_{1leq ileq n } J_i}simeq R/J_1oplus ...oplus R/J_n.$$
In the proof the fact that the ideals are bilateral is used not only to make sense of the ring quotient $R/J_i$.
So I’m wondering whether I could extend the theorem in order to obtain a module isomorphism instead of a ring one in the following sense:

Let R be a non-commutative ring (the commutative case is in fact the Chinese remainder Theorem itself) and $J_1,dots , J_nsubset R$ be left ideals such that $J_i+J_k=R$ if $ineq k$. Let $R/J_i$ be the quotient module of the left regular R-module modulo the left $R$-module $ J_i$. Then the mapping $varphi$ (defined above) is a surjective R-module morphism.

The problem rose in trying to prove (or disprove) that any finite dimensional $k$-algebra has only a finite number of simple left modules up to isomorphism. If the module-generalisation theorem holds, then I have proved this fact.

The answer that follows is just a detailed version of my comments to the
original post.

Your conjecture (that $varphi$ is surjective) is true for each $nleq2$ but
false for each $ngeq3$. Let me show this. First, let me prove that it is true
for $n=2$ (for the sake of completeness -- I know that you have a proof):

Proposition 1. Let $R$ be a (noncommutative, associative, unital) ring.
Let $U$ and $V$ be two left ideals of $R$ such that $U+V=R$. Let
$varphi:Rrightarrowleft( R/Uright) oplusleft( R/Vright) $ be the
map that sends each $rin R$ to $left( r+U,r+Vright) inleft(
R/Uright) oplusleft( R/Vright) $. Then, $varphi$ is a surjective
$R$-module homomorphism.

Proof of Proposition 1. It is clear that $varphi$ is an $R$-module
homomorphism. It thus remains to prove that $varphi$ is surjective.

We have $1in R=U+V$. In other words, there exist $uin U$ and $vin V$ such
that $1=u+v$. Consider these $u$ and $v$.

Let $zinleft( R/Uright) oplusleft( R/Vright) $ be arbitrary. Thus,
we can write $z$ in the form $z=left( alpha,betaright) $ for some
$alphain R/U$ and $betain R/V$. Consider these $alpha$ and $beta$.

We have $alphain R/U$; thus, there exists some $ain R$ such that
$alpha=a+U$. Consider this $a$.

We have $betain R/V$; thus, there exists some $bin R$ such that $beta
=b+V$. Consider this $b$.

Let $x=av+buin R$. Thus,
begin{align*}
underbrace{x}_{=av+bu}-underbrace{a}_{substack{=a1=aleft( u+vright)
\text{(since }1=u+vtext{)}}} & =av+bu-aleft( u+vright) =bu-au=left(
b-aright) underbrace{u}_{in U}\
& inleft( b-aright) Usubseteq Uqquadleft( text{since }Utext{ is a
left ideal of }Rright) ,
end{align*}
so that $x+U=a+U=alpha$. Also,
begin{align*}
underbrace{x}_{=av+bu}-underbrace{b}_{substack{=b1=bleft( u+vright)
\text{(since }1=u+vtext{)}}} & =av+bu-bleft( u+vright) =av-bv=left(
a-bright) underbrace{v}_{in V}\
& inleft( a-bright) Vsubseteq Vqquadleft( text{since }Vtext{ is a
left ideal of }Rright) ,
end{align*}
so that $x+V=b+V=beta$. Now, the definition of $varphi$ yields
begin{equation}
varphileft( xright) =left( underbrace{x+U}_{=alpha},underbrace{x+V}
_{=beta}right) =left( alpha,betaright) =z.
end{equation}
Thus, $z=varphileft( underbrace{x}_{in R}right) invarphileft(
Rright) $.

Now, forget that we fixed $z$. Thus, we have shown that $zinvarphileft(
Rright) $ for each $zinleft( R/Uright) oplusleft( R/Vright) $. In
other words, the map $varphi$ is surjective. This completes the proof of
Proposition 1. $blacksquare$

Proposition 1 shows that your conjecture holds for $n=2$. For $n<2$, it is
completely obvious (since $varphi$ is a projection map in this case). Now,
let me disprove your conjecture for $n>2$ using the following example:

Example 2. Let $n>2$ be an integer. Let $V$ be the $2$-dimensional
$mathbb{Q}$-vector space $mathbb{Q}^{2}$. For each positive integer $i$, let
$e_{i}$ be the vector $left( 1,iright) ^{T}in V$. Note that these vectors
$e_{1},e_{2},e_{3},ldots$ are pairwise linearly independent. Let
$R=operatorname*{End}nolimits_{mathbb{Q}}Vcongmathbb{Q}^{2times2}$. For
each positive integer $i$, let $J_{i}$ be the subset $left{ Ain
R mid Ae_{i}=0right} $ of $R$. It is clear that all these subsets
$J_{1},J_{2},J_{3},ldots$ are left ideals of $R$. Moreover, each $J_{i}$ is a
$2$-dimensional subspace of the $4$-dimensional $mathbb{Q}$-vector space $R$.
But any two positive integers $i$ and $j$ satisfy $J_{i}cap J_{j}=0$ (because
any endomorphism $Ain R$ that annihilates the two linearly independent
vectors $e_{i}$ and $e_{j}$ must be the zero map) and therefore $J_{i}
+J_{j}=R$ (since $dimleft( J_{i}+J_{j}right) =underbrace{dimleft(
J_{i}right) }_{=2}+underbrace{dimleft( J_{j}right) }_{=2}
-underbrace{dimleft( J_{i}cap J_{j}right) }_{=0}=4$). Hence, if your
conjecture were true, the map $varphi:Rrightarrowleft( R/J_{1}right)
oplusleft( R/J_{2}right) opluscdotsoplusleft( R/J_{n}right) $
would be surjective. This would yield $4geq2n$, since this map $varphi$ is a
$mathbb{Q}$-linear map from a $4$-dimensional $mathbb{Q}$-vector space to a
$left( 2nright) $-dimensional $mathbb{Q}$-vector space; but this would
contradict $n>2$.

Finally, the question you were trying to solve in the last paragraph is a
consequence of Theorem 1 in https://mathoverflow.net/a/14516/ .