Let $R$ be a ring. The Chinese Remainder Theorem asserts that if $J_1,\ \ldots,\ J_n\lhd R$ are bilateral coprime ideals ( i.e. $J_i+J_k=R,\ \forall i\neq k$) then the map $$\varphi:a\in R\longmapsto (a+J_1,...,a+J_n)\in R/J_1\oplus ...\oplus R/J_n $$ is a surjective ring morphism, such that $$\ker (\varphi)=\displaystyle \bigcap _{1\leq i\leq n} J_i.$$ Thus we have a ring isomorphism (and thus a R-module isomorphism) $$\frac{R}{\displaystyle \bigcap_{1\leq i\leq n } J_i}\simeq R/J_1\oplus ...\oplus R/J_n.$$ 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 $J_1,\dots ,\ J_n\subset R$ be left ideals such that $J_i+J_k=R$ if $i\neq k$. 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.

  • $\begingroup$ Interesting question. It is certainly true for $n = 2$. $\endgroup$ – darij grinberg Nov 20 '18 at 20:24
  • $\begingroup$ I’ve already noted this since the proof for the bilateral case still holds, but I forgot to put it in the post. Thank you $\endgroup$ – Blumer Nov 20 '18 at 20:26
  • $\begingroup$ Ah, it is false for $n = 3$. Let $V$ be a $2$-dimensional $\mathbb{Q}$-vector space with basis $\left(e_1, e_2\right)$. Let $e_3 = e_1 + e_2$. Let $R = \operatorname{End} V$. For each $i$, let $J_i$ be the left ideal $\left\{ A \in R \mid A e_i = 0 \right\}$. Then, $J_i + J_j = R$ whenever $i \neq j$, but the map $\varphi$ cannot be surjective (since it goes from a $4$-dimensional $\mathbb{Q}$-vector space into a $6$-dimensional one). $\endgroup$ – darij grinberg Nov 20 '18 at 20:33
  • $\begingroup$ As for the question you were trying to solve in the last paragraph: It follows from Theorem 1 in mathoverflow.net/questions/14514/… . $\endgroup$ – darij grinberg Nov 20 '18 at 20:34
  • $\begingroup$ @darijgrinberg sorry but I can’t see how this is related with my latter question: I suppose that there exist k-algebras which are not group algebras for any group. $\endgroup$ – Blumer Nov 21 '18 at 21:45

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 $n\leq2$ but false for each $n\geq3$. 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:R\rightarrow\left( R/U\right) \oplus\left( R/V\right) $ be the map that sends each $r\in R$ to $\left( r+U,r+V\right) \in\left( R/U\right) \oplus\left( R/V\right) $. 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 $1\in R=U+V$. In other words, there exist $u\in U$ and $v\in V$ such that $1=u+v$. Consider these $u$ and $v$.

Let $z\in\left( R/U\right) \oplus\left( R/V\right) $ be arbitrary. Thus, we can write $z$ in the form $z=\left( \alpha,\beta\right) $ for some $\alpha\in R/U$ and $\beta\in R/V$. Consider these $\alpha$ and $\beta$.

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

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

Let $x=av+bu\in R$. Thus, \begin{align*} \underbrace{x}_{=av+bu}-\underbrace{a}_{\substack{=a1=a\left( u+v\right) \\\text{(since }1=u+v\text{)}}} & =av+bu-a\left( u+v\right) =bu-au=\left( b-a\right) \underbrace{u}_{\in U}\\ & \in\left( b-a\right) U\subseteq U\qquad\left( \text{since }U\text{ is a left ideal of }R\right) , \end{align*} so that $x+U=a+U=\alpha$. Also, \begin{align*} \underbrace{x}_{=av+bu}-\underbrace{b}_{\substack{=b1=b\left( u+v\right) \\\text{(since }1=u+v\text{)}}} & =av+bu-b\left( u+v\right) =av-bv=\left( a-b\right) \underbrace{v}_{\in V}\\ & \in\left( a-b\right) V\subseteq V\qquad\left( \text{since }V\text{ is a left ideal of }R\right) , \end{align*} so that $x+V=b+V=\beta$. Now, the definition of $\varphi$ yields \begin{equation} \varphi\left( x\right) =\left( \underbrace{x+U}_{=\alpha},\underbrace{x+V} _{=\beta}\right) =\left( \alpha,\beta\right) =z. \end{equation} Thus, $z=\varphi\left( \underbrace{x}_{\in R}\right) \in\varphi\left( R\right) $.

Now, forget that we fixed $z$. Thus, we have shown that $z\in\varphi\left( R\right) $ for each $z\in\left( R/U\right) \oplus\left( R/V\right) $. 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 $\left( 1,i\right) ^{T}\in V$. Note that these vectors $e_{1},e_{2},e_{3},\ldots$ are pairwise linearly independent. Let $R=\operatorname*{End}\nolimits_{\mathbb{Q}}V\cong\mathbb{Q}^{2\times2}$. For each positive integer $i$, let $J_{i}$ be the subset $\left\{ A\in R\ \mid\ Ae_{i}=0\right\} $ 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 $A\in R$ 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 $\dim\left( J_{i}+J_{j}\right) =\underbrace{\dim\left( J_{i}\right) }_{=2}+\underbrace{\dim\left( J_{j}\right) }_{=2} -\underbrace{\dim\left( J_{i}\cap J_{j}\right) }_{=0}=4$). Hence, if your conjecture were true, the map $\varphi:R\rightarrow\left( R/J_{1}\right) \oplus\left( R/J_{2}\right) \oplus\cdots\oplus\left( R/J_{n}\right) $ would be surjective. This would yield $4\geq2n$, since this map $\varphi$ is a $\mathbb{Q}$-linear map from a $4$-dimensional $\mathbb{Q}$-vector space to a $\left( 2n\right) $-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/ .


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