let I,J be nonzero ideal in a domain R,show that I $\cap$ J$\neq {0}$
my attempt,for any two ideals,there is a isomorphic \begin{equation*} \frac{I}{(I\cap J)}\cong \frac{(I+J)}{J} \end{equation*} so,if $I\cap J={0}$,then $I\cong \frac{(I+J)}{J}$ by $a\rightarrow a+J$.How can I get a contradiction.
Note that $I \cap J$ is never empty, since every ideal contains $0$. What is wanted here is $I \cap J \ne \{0\}$, the trivial or zero ideal, thusly:
I assume by "domain" our OP means "integral domain"; then $R$ is a commutative ring such that
$\forall 0 \ne x, y \in R, \; xy \ne 0; \tag 1$
this of course yields
$xy = 0 \Longrightarrow [x = 0] \vee [y = 0]; \tag 2$
that is, a product is only zero when one of the factors vanishes.
Now with $I$, $J$ non-trivial ideals such that
$I \cap J = \{0\}, \tag 3$
then since
$0 \ne i \in I, 0 \ne j \in J \Longrightarrow ij \in I \cap J, \tag 4$
in accord with (3)
$ij = 0; \tag 5$
but by (2) this implies at least one of $i$, $j$ vanishes, contradicting (4). Thus (3) is false and
$I \cap J \ne \{0\}. \tag 6$