Let $I$ and $J$ be ideals in $R$. Is the set $K= \{ ab \ | \ a\in I, b\in J \}$ an ideal in R? [duplicate]

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I've just assumed that this is false, since the problem statements says to compare it to a previous problem where $\{ a+b \ | \ a\in I, b\in J \}$ is ideal.

However, by trial and error I can't find two ideals where this doesn't hold.

Is this false, and if so what's the counterexample? I haven't found one thus far.

marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 18 '16 at 10:35

• this works if we take $(x,y)$ twice and consider $x^2+y^2$. – Jorge Fernández Hidalgo Mar 17 '16 at 23:20
• Or $I=J=\langle 2,X\rangle$ and consider $4+X^2$. – Henning Makholm Mar 17 '16 at 23:24