If $I$ and $J$ are Ideals of a commutative ring $R$ where $I + J = R$, then $c \equiv a \pmod I$ and $c \equiv b \pmod J$.

Question

Let $R$ be a commutative ring with unity. Let $I$ and $J$ be Ideals of $R$ and let $I + J = R$. I want to show that if $a,b \in R$, then there exists some $c \in R$ such that $c \equiv a \pmod I$ and $c \equiv b \pmod J$.

I am having trouble even starting this problem. The only thing I have to start with is that if $a$ and $b$ are in $R$ then $a = x_1 + y_1$ and $b= x_2 + y_2$ where $x_1,x_2 \in I$ and $y_1,y_2 \in J$. Any help would be appreciated.

We know that $1 = x+y$ for some $x \in I$ and $y \in J$. Now consider $c := ay+bx$. — azif00, Jul 18 '21 at 22:31
I agree that 1 can be written as such and I am not entirely sure about your intuition on how you defined $c$ but my biggest concern is the significance $\mod I$ and $\mod J$ in the context of this problem? — An Isomorphic Teen, Jul 18 '21 at 22:40

score 2 · Answer 1 · answered Jul 18 '21 at 22:58

We have that $1 = x + y$ for some $x \in I$, $y \in J$. Let $a,b \in R$ and define $c \in R$ such that $c = ay + bx$. Now $1 = x + y \implies y = 1 - x$ and $x = 1 - y$. Thus we have that $$c = ay + bx = a(1 - x) +bx.$$ This then implies that $$c - a = bx - ax.$$ Now as $I$ is an ideal, $bx - ax \in I$. Hence $c - a \in I$ and thus $c \equiv a \mod I.$ A similar argument shows that $c \equiv b \mod J$.

Bill Dubuque · Answer 2 · 2021-07-19T00:41:13.280

It's simply special case $\,\color{#c00}{I + J = (1)}\,$ in the ideal form of the CRT solvability criterion below, which follows immediately from the proof in the classical case $\,R = \Bbb Z,\,$ i.e. the proof boils down to a trivial Equation Rearrangement $\rm \color{darkorange}{ER}$ of the Bezout equation.

Theorem $\ \ \ \ \ \left.\exists\, x\in R\!:\,\ \begin{align}x\equiv a\!\!\pmod{\!I}\\ x\equiv \color{#0a0}{b\!\!\pmod{\!J}}\end{align}\,\right\} \,\iff\,\ \ a\!-b \in \color{#c00}{I+J}$

Proof $\ \ \exists\, i,j\in I,J\!:\ a\! +\! i = \color{#0a0}{b\! +\! j}\!\!\underset{\large \rm\color{darkorange}{ER}\!\!\!}\iff\! \exists\, i,j\in I,J\!:\,\ a\!-\!b= \color{#c00}{i-j} $

score 1 · Answer 3 · answered Nov 13 '21 at 08:04

This is a natural consequence of the Chinese Remainder Theorem for Rings (with ring $\mathbb Z$ and coprime ideals $I = m \mathbb Z$ and $J = n \mathbb Z$ for coprime $m$ and $n$ as a special case): $R / IJ \cong R/I \times R/J $.

The proof, available many places online, is to conclude our natural isomorphism $r + IJ \mapsto (r+I,r+J) $ using natural ring homomorphism $\varphi: R \to R/I \times R/J $ defined by $\varphi(r) = (r+I, r+J)$ and the First Isomorphism Tehorem.

Just as Bézout's identity says there exists $x,y \in \mathbb Z$ with $mx + ny = 1$, we define coprime ideals as $I + J = (1) = R$. This lets us show things like $I \cap J = IJ$ and $\varphi$ is surjective.

If $I$ and $J$ are Ideals of a commutative ring $R$ where $I + J = R$, then $c \equiv a \pmod I$ and $c \equiv b \pmod J$.

3 Answers3

Linked