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Let $A$ and $B$ be algebras over a field $K$. The ideals of the tensor product $A\bigotimes_K B$ are of the form $I\bigotimes_K J$ where $I$ is an ideal of $A$ and $J$ is an ideal of $B$?

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    $\begingroup$ No. The polynomial ring $K\left[X,Y\right]\cong K\left[X\right]\otimes K\left[Y\right]$ over a field $K$ should give you a good hint about how complicated the ideals of a tensor product can get. $\endgroup$ Jan 20, 2012 at 15:28
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    $\begingroup$ For an even more complicated situation, consider $K[[x]]\otimes K[[y]]$, a tensor product of two noetherian rings which results in a non-noetherian ring. $\endgroup$
    – the L
    Jan 20, 2012 at 15:30
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    $\begingroup$ Let $K|k$ be a finite extension of fields. Comparing $k$-dimensions shows that multiplication $K \otimes_k K \to K$ must have non-trivial kernel which thus cannot be of the form you expected. $\endgroup$
    – Ralph
    Jan 20, 2012 at 15:38
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    $\begingroup$ Note to people downvoting the question: it is not a completely wild guess. See, e. g., Lemma 2.7 in Chapter IV of Milne's Class Field Theory ( jmilne.org/math/CourseNotes/cft.html ) for a case when it is true. $\endgroup$ Jan 20, 2012 at 15:38
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    $\begingroup$ Well I don't think that the questioner has considered any examples before posting this. See also the FAQ mathoverflow.net/faq. -1 $\endgroup$ Jan 20, 2012 at 16:16

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