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Let $R:=\mathbb{Z}_2[w_1,\ldots,w_n]$, $I$ some ideal in $R$ and $A=R/I$. I would very much want to show that a sequence of the form:

$0\to A/(\alpha)\overset{\cdot\beta}{\to}A/(\alpha\beta)\overset{id}{\to}A/(\beta)\to 0$

is a SES of $R$ Modules.

In order to prove the injectivity of the first homomorphism (which is the tricky part), it would suffice to show that $(\beta)\cap I=(\beta)\cdot I$. Thus, the main question is if there exists a well documented notion for two ideals $I,J$ to satisfy $I\cap J$=$I\cdot J$.

Of course some cases are known. If the ideals are coprime for example, or if they are both radical, the desired property follows. I have found the very same question here, but since that question was motivated by algebraic-geometry, the answer about the $Tor$ functor seems to have covered the topic.

Does this property even have a name? Do there exist closure results like: If $I=(a)$ and $J=(b_1,b_2)$ have this property, then $I$ and $J'=(b_1,rb_2+r'a)$ have this property as well (for some good choices of $r,r'$). (which incidentally is a lemma that would solve my SES problem, due to the exact choice of $I$ and $\beta$)

If the answer to all these is just "only what you've already found in the link seems to be documented", then my follow-up question is if someone has an idea when the sequence in the beginning is exact and how I could proceed, without messing around with the above mentioned property. (I have tried to go through localizations for example, but since I am not that familiar with the topic, I quickly got stuck)

If it is needed, I could elaborate on the choice of $I$,$\alpha$ and $\beta$.

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  • $\begingroup$ The one situation where this is obvious is if $\alpha\beta$ is a non-zero divisor in $A$. For injectivity on the left, you need just that $\beta$ is a non-zero divisor in $A$. If that is not the case, in general this can be false without further conditions. $\endgroup$
    – Mohan
    Nov 1, 2017 at 1:04

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