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Let $R$ be a commutative ring with $1$. If $I+J=R$, then $IJ = I \cap J$. The post below has already given a solution. However, I am wondering what happens if $R$ is not commutative? Can anyone provide me with a counterexample? If I am not wrong, the counterexample given in the post is the case when $R$ is commutative and does not have unity. Which part of the proof uses commutativity of the ring? Thank you.

If $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.

user26857
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user10024395
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  • See also https://math.stackexchange.com/questions/3376478/chinese-remainder-theorem-does-not-hold-in-non-commutative-case . – darij grinberg Jan 22 '21 at 03:10

1 Answers1

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It suffices to find ideals $I,J$ such that $I+J=R$ and $IJ \neq JI$. In fact, then $IJ = I \cap J$ and $JI = J \cap I$ cannot hold both, so that either $(I,J)$ or $(J,I)$ is a counterexample, or both, as below.

Consider $R=\mathbb{Z}\langle x,y \rangle / (2x+2y=1)$, $I=(x)$, $J=(y)$. Then $I+J=R$, and one can prove that $I \cap J = (xy,yx)$, which strictly contains $IJ=(xy)$. (In order to simplify calculations, you may work with $R/(xy)$.)

Here is a positive result:

If $I,J$ are two-sided ideals of a ring $R$ with $I+J=R$, then $IJ+JI = I \cap J$.

Proof: The inclusion $IJ +JI \subseteq I \cap J$ follows from $IJ \subseteq I$, $IJ \subseteq J$, $JI \subseteq I$, $JI \subseteq J$. Conversely, $I \cap J = (I \cap J)R = (I \cap J)(I+J) = (I \cap J)I + (I \cap J)J \subseteq JI+IJ$. $\checkmark$

Martin Brandenburg
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  • Where does the "strictly" in your counterexample come from? I seem to get $IJ = I \cap J$ (and in fact, $xy = yx$). Indeed, $2x + 2y = 1$ shows that $2x = 1 - 2y$, whence $2x$ commutes with $y$. Thus, $2x$ is central in $R$. Likewise, $2y$ is central in $R$. Hence, $2xy$ is central in $R$ (because any $z \in R$ satisfies $(2xy)z = x(2y)z = xz(2y) = (2x)zy = z(2x)y = z(2xy)$), and similarly any product of the form $2 \cdot\left(\text{any monomial in } x \text{ and } y\right)$ is central in $R$. But $x$ is a sum of two such monomials, since $x = x\left(2x+2y\right) = 2xx + 2xy$. – darij grinberg Jan 22 '21 at 02:59
  • Easier argument for why $R$ is secretly commutative: By a linear substitution of variable ($z = x+y$), rewrite the definition of $R$ as $R \cong \mathbb Z\left<z,x\right> / \left(2z = 1\right) \cong \mathbb{Z}\left[1/2\right]\left<x\right> \cong \mathbb{Z}\left[1/2\right]\left[x\right]$. – darij grinberg Jan 22 '21 at 03:09
  • Here is an actual counterexample: https://math.stackexchange.com/questions/3376478/chinese-remainder-theorem-does-not-hold-in-non-commutative-case – darij grinberg Jan 22 '21 at 03:13