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Let $R$ be an integral domain, and $A , B$ be non trivial ideals of $R$. Then prove that $|A \cap B|>1$.

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Let $0 \ne a \in A$, $0 \ne b \in B$; we know such $a, b$ exist by virtue of the assumption that $A$ and $B$ are nontrivial. Then since $A$ and $B$ are ideals of the integral domain $R$, we have $0 \ne ab \in A$ and $0 \ne ab \in B$, whence $ab \in A \cap B$. Furthermore, for any $r \in R$, we have $abr \in A$ and $abr \in B$, so that in fact $abr \in A \cap B$, or $(ab) = abR \subset A \cap B$. Now for nonzero $s \in R$ the map $\phi_s: R \to R$ defined by $\phi_s(c) = sc$ is clearly injective, for if $\phi_s(c_1) = \phi_s(c_2)$ then $sc_1 = sc_2$, so by the cancellation property of integral domains we have $c_1 = c_2$. Also, we see that $\phi_{ab}(R) = abR \subset A \cap B$. Thus $A \cap B$ has a subset, $abR$, with $\mid abR \mid = \mid R \mid$. Since nontrivial integral domains $R$ must have $0 \ne 1$, $\mid R \mid \ge 2$, whence $\mid abR \mid \ge 2$. Since $abR \subset A \cap B$, we have $\mid A \cap B \mid \ge 2$; in fact, we have shown that $\mid A \cap B \mid \ge \mid R \mid$; but since $A \cap B \subset R$, we indeed obtain $\mid A \cap B \mid = \mid R \mid$ as well. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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    $\begingroup$ Dear RobertLewis: everything after "Furthermore" is superfluous. Before "Furthermore" you've already shown $A\cap B$ contains an element other than $0$, so already $|A\cap B|>1$. Please consider striking out the rest: it might be confounding to readers. Regards. $\endgroup$ – rschwieb Dec 15 '13 at 14:22
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We know that $0\in A$ and $0\in B$. Let $a\in A$ and $b\in B$. Then $ab\in A\cap B$. Since $R$ is a integral domain, $ab\neq 0$.

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