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By irreducible ideal I of a ring A, I mean:

I is irreducible if $\forall b,c$ ideals in A that satisfy $I= b \cap c$, then $I=b$ or $I=c$.

I managed to prove the implication from left to right, but have trouble proving right to left. My attempt:

Let $\varphi :A \longrightarrow$ A/I be the canonic surjection. Let $b,c\leqslant A$ such that $I= b \cap c$.

Then, $\varphi(I) = \varphi(b \cap c)= (\bar{0}) \subseteq \varphi(b) \cap \varphi(c)$ and I know that the image of intersection is not the intersection of images unless $\varphi$ is injective (which is not).

This is my first post in this forum, where can I learn more Latex symbols for algebra?

Mejay
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    It is called an "irreducible" ideal. And here is a tutorial for MathJax. – Dietrich Burde Oct 04 '23 at 18:43
  • Actually, the intersection is equal to the intersection of images. This is true when $b,c$ are ideals containing $I$, not when they are arbitrary subsets. So this doesn't contradict the non-injectivity of $\varphi$. Assume we have an element in $\varphi(b)\cap\varphi(c)$. Then it is of the form $x+I=y+I$ where $x\in b, y\in c$. Then $x-y\in I=b\cap c$. Hence $x=y+(x-y)\in c$, as a sum of two elements of $c$. So $x\in b\cap c$, which implies $x+I\in\varphi(b\cap c)$. – Mark Oct 04 '23 at 18:50
  • @DietrichBurde Thank you,my romanian book names it as in "irreductible. – Mejay Oct 04 '23 at 19:20
  • @Mark Thanks, I will try to continue my proof knowing φ(b)∩φ(c)=φ(b∩c). – Mejay Oct 04 '23 at 19:21

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Note that this is a very short proof using Correspondence. That is, let $\pi:A\to A/I$ be defined by $\pi(a)=a\pmod{I}$. Then $J, K$ are ideals in $A/I$ with $J\cap K=(0)$ if and only if $\pi^{-1}(J)$ and $\pi^{-1}(K)$ are ideals in $A$ with $\pi^{-1}(J)\cap \pi^{-1}(K)=I$.

IAAW
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