# Quasi-Chinese remainder theorem for modules

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.

• Interesting question. It is certainly true for $n = 2$. – darij grinberg Nov 20 '18 at 20:24
• 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 – Blumer Nov 20 '18 at 20:26
• 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). – darij grinberg Nov 20 '18 at 20:33
• As for the question you were trying to solve in the last paragraph: It follows from Theorem 1 in mathoverflow.net/questions/14514/… . – darij grinberg Nov 20 '18 at 20:34
• @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. – Blumer Nov 21 '18 at 21:45

## 1 Answer

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 $$$$\varphi\left( x\right) =\left( \underbrace{x+U}_{=\alpha},\underbrace{x+V} _{=\beta}\right) =\left( \alpha,\beta\right) =z.$$$$ Thus, $$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/ .