Understanding surjectivity in Chinese Remainder Theorem Proof

Question

Im self-studying some ring theory and in Dummit and Foote Chapter 7.6 Theorem 17, pg 265-266 they have a proof for the case where $k = 2$ that I am struggling to understand--maybe I'm forgetting some identity or don't have the correct grasp on 'mod'-- but basically it says that if $A,B $ are ideals of $R$, the map $\phi : R \to R/A \times R/B$ defined by $\phi(r) = (r + A, r+ B)$ is a surjective homomorphism iff A and B are comaximal. I see that the function is a homomorphism but dont follow their reasoning as to why its surjective. I understand the notation $r+A \in R/A$ far more than the mod notation, if that helps.

Answer 1

Suppose $\phi$ is surjective. Then for every $x,y$ there exists $ z$ such that $x+A=z+A, y+B=z+B$ so $ z-x\in A, z-y\in B$. Take $x=1,y=0$. So there exists $z$ such that $z-1\in A, z\in B$. Hence $A+B$ contains $1$, $A+B=R$, so $A,B$ are co-maximal.

Now suppose that $A,B$ are co-maximal, $a+b=1$ where $a\in A, b\in B$. Then take any $x,y\in R$. We need to find a $z\in R$ such that $z-x\in A, z-y\in B$. So we need $x+a'=y+b'$ for some $a'\in A, b'\in B$, or $x-y=b'-a'$. But $x-y=(x-y)\cdot 1$, so $x-y=(x-y)(a+b)= (x-y)*b-(x-y)(-a)$. Take $z=x+(x-y)(-a)=x(1-a)+ya=xb+ya=xb+y-yb.$ Hence $z\in x+A, z\in y+B$.

You forgot to include one part of the theorem: that if $A$, $B$ are co-maximal, then $A\cap B=AB$. That is also easy. Clearly, $AB\subseteq A\cap B$. Suppose $a+b=1$, $a\in A, b\in B$ and $z\in A\cap B$. Then $z=z\cdot 1= za+zb\in BA+AB=AB$ since the ring is commutative.

Answer 2

The idea of the surjection part of the proof is clearer using vector language in the product ring, which has pointwise ring operations, e.g. $\,(a,b,c)(a',b',c') = (aa',bb',cc')$. In particular the colored vectors below vividly show how the $\color{#0a0}{n\text{-ary}}$ idempotents $\,e_i\,$ are constructed simply by employing products of idempotents from $\rm\color{#0a0}{binary}$ subproducts $A_i\!\times\! A_j$ via $\,A_i\!+\!A_j\!=\!(1),\,i\!\neq\! j$.

By linearity, it is clear that $\,\phi$ is onto if it hits the "basis" vectors (complete idempotent system) $\,e_1\mapsto (1,0,0,\ldots),\,$ $\, e_2\mapsto (0,1,0,\ldots),\,\ldots,\,$ since then we have e.g. $$\begin {align} &a_1 e_1 + a_2 e_2 + a_3 e_3\\[.3em] \mapsto\ &\bar a_1(1,0,0)+\bar a_2(0,1,0)+\bar a_3(0,0,1)\\[.3em] =\,\ & (\bar a_1, \bar a_2, \bar a_3) \end{align}\qquad$$

To construct the $\,e_i\,$ we apply the pair-coprime hypotheses $A_j+A_k = (1)$ if $\,j\neq k\,$ as below.

$\ \ \ \ \begin{align} &\!\!\!\!\!\!\!\!\!\!\!\!\text{ $\ \ \ \ e_1 \mapsto (1,\color{#0a0}0,\color{#c00}0,\ldots,\color{darkorange}0)\,$ is the $\textit{product}$ of the following $\,v_j\,$ ($*\,$ denotes any element)}\\ v_2 &\mapsto\overbrace{(1^{\phantom{.}\!\!},\color{#0a0}0,*,*,\ldots)},\ \ {\rm i.e.}\ \ v_2 \equiv 1\!\!\!\pmod{\!\!A_1},\,\ v_2\equiv 0\!\!\!\pmod{\!\!A_2};\,\ \color{#0af}{\text{all $v_j$ exist}}\ \rm by\ below\\ v_3 &\mapsto (1,*,\color{#c00}0,*,\ldots),\ \ {\rm i.e.}\ \ v_3 \equiv 1\!\!\!\pmod{\!\!A_1},\,\ v_3\equiv 0\!\!\!\pmod{\!\!A_3}\\ v_4 &\mapsto (1,*,*,0,\ldots),\ \ \cdots\\[-.4em] &\ \ \vdots\qquad\qquad\qquad\quad\ \ \: {\rm i.e.}\ \ v_j\equiv 1\!\!\!\pmod{\!\!A_1},\,\ v_j\equiv 0\!\!\!\pmod{\!\!A_j}\\[-.4em] v_k &\mapsto (1,*,*,\ldots,\color{darkorange}0) \end{align}$

$\color{#0af}{\text{All $v_j$ exist}}$ since by hypothesis $\,A_1+A_j = (1)\,$ so $\, a_1 + v_j = 1\,$ for some $\,a_1\in A_1,\, v_j\in A_j\,$ so $\, v_j = 1-a_1\equiv 1\pmod{\!\!A_1}\,$ by $\,a_1\in A_1,\,$ and $\,v_j\equiv 0\pmod{\!\!A_j}\,$ by $\,v_j\in A_j,\,$ as sought.

We get all other basis vectors $\,e_2 = (0,1,0,\ldots),\, e_3 = (0,0,1,\ldots),\ldots\,$exactly the same way.

Remark $ $ In a PID $\,A_j = (a_j)\,$ so we can choose $\,v_j = a_j({a_j}^{-1}\bmod a_1),\,$ so the above product is $\,e_1 = v_2\cdots v_k = m_1({m_1}^{-1}\bmod a_1),\ m_1 = a_2 a_3\cdots a_k,\,$ yielding the classical CRT formula.

Also we have $\,(1) = A_i + A_j =(a_i)+(a_j) = (\gcd(a_i,a_j))\,$ so this hypothesis means that the moduli $\,a_i\,$ are pairwise coprime - the classical CRT solvability criterion. As explained there this generalizes widely (it is one of many known characterizations of Prufer domains - non-Noetherian generalizations of PIDs and Dedekind domains).