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Suppose $R$ is a commutative and unital ring. Let the ideal $I$ be maximal and $a,b$ be (nonzero) zero divisors in $R$.

Show that $ab = 0$ implies $a \in I$ or $b\in I$

We've only had a bit of exposure to ideals: we know that $I$ maximal $\to R/I$ field, a little about the Euclidean algorithm, and the definition of a PID.

I'm not sure how to approach this. The problem seems simple and I'm probably just missing something.

Should I try assuming $a,b \notin I$ and try to derive a contradiction?

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5 Answers 5

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For maximal ideals $I\subset R$, it can be checked that the quotient $R/I$ is a field. (It is actually an equivalent condition). If $a$ is not in $I$, then $a+I\in R/I$ is a non-zero element and hence has an inverse ($R/I$ being a field). Sameway if $b$ is not in $I$, then $b+I$ also has an inverse in $R/I$. Then the product of those inverses will be the inverse for $ab+I$, hence $ab$ is also not in $I$.

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Hint: If $ab = 0$, then $ab+I = I$. Notice that $ab+I=(a+I)(b+I)$.

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If we work modulo $I$, then $[a] \cdot [b] = [a \cdot b] = [0]$, where $[a],[b]$ denote the classes of $a,b$ modulo $I$. As $R/I$ is a field, it follows that $[a] = 0$ or $[b] = 0$ which implies that $a \in I$ or $b \in I$.

If you know that maximal ideals are prime, then this also follows more or less immediately from the definition of prime ideal.

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In a commutative ring $R$, a maximal ideal $A$ is prime. $A$ maximal $\Rightarrow R/A$ is a field $\Rightarrow R/A$ is an integral domain $\Rightarrow A$ is prime.

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If $I$ is a maximal ideal then $I$ is a prime ideal. Indeed, let $ab \in I$ such that $a \notin I$. It suffices to show $b \in I$. Now, $I$ is a subring of $I+Ra$ and $I+Ra$ is an ideal, so $R=I+Ra$. Then there exists $m \in I$ and $r \in R$ satisfying $1=m+ra$. Thus, $b=bm+rab \in I+I \subset I$.

Now, the claim is that then $R$ is an integral domain if and only if $(0)$ is a prime ideal

Indeed, $(0)$ is a prime ideal if and only if $ab \in (0)$ implies $a \in (0)$ or $b \in (0)$ if and only if $ab=0$ implies $a=0$ or $b=0$.

Here, we have shown that if $I$ is maximal, then $I$ is a prime ideal, and further take $I=(0)=0.$ The result should now follow.

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