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Let $K$ be a separable $F$-algebra, and $E$ be a extension field of $F$. We see that $E\otimes _FK$ is a separable $E$-algebra. Then $E\otimes _FK$ is a separable extension field of $E$.

$E\otimes _FK$, as a field, is a simple ring.

However, $E\otimes _FK$, as a separable $E$-algebra, is a direct sum of some simple $E$-algebras, which we denote by $E_i$. Then $E\otimes _FK$ direct sum of separable extension fields $E_i$ of $E$ (It is well known that a separable $E$-algebra is a separable extension field over $E$). Then $E\otimes _FK=E_1\oplus\cdots \oplus E_r$, which have nonzero ideals $E_1,\cdots,E_r)$...

It may be looks like a contraction.

Thanks everyone!

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    The contradiction disappears when you conclude that $r=1$ (indeed a simple ring has a unique direct summand!).. – 2017-01-12

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It is simply incorrect that $E\otimes_F K$ is a field. This is true sometimes (in the case that you end up having $r=1$), but not in all cases. For instance, $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong\mathbb{C}\times\mathbb{C}$, which is not a field. See https://mathoverflow.net/questions/82083/when-is-the-tensor-product-of-two-fields-a-field for a discussion of when the tensor product of two fields is a field.

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    Let $K$ be a separable $F$-algebra, and $E$ be a extension field of $F$. We see that $E\otimes _FK$ is a separable $E$-algebra. It is a well known result that $E\otimes _FK$ is a separable extension field of $E$. – 2017-01-14
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    That "well-known result" is just false. I'm guessing you're misreading something somewhere--do you have a source you are getting that result from? – 2017-01-14
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    Proposition 7.4 in the book "Methods of representation Theory(I)" from Charles W. Curtis... – 2017-01-15
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    I add the image concerning the question above, thank you! – 2017-01-15
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    Notice that that proposition begins by assuming that $E$ is a finite extension _field_ of $K$. If you don't already know $E$ is a field, it doesn't tell you anything. – 2017-01-15
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    In the question I have assume that $E$ is a finite extension field of $K$.. – 2017-01-15
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    But you're trying to apply the proposition to $E\otimes_F K$, not to the $E$ of the question. – 2017-01-15
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    Why $E⊗_FK$ is not a field? – 2017-01-15
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    I gave an example where it is not a field in my answer. A quick way to see $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}$ is not a field is to note that $(i\otimes 1+1\otimes i)(i\otimes 1-1\otimes i)=i^2\otimes 1-1\otimes i^2=-1\otimes 1+1\otimes 1=0$, so it has zero divisors. – 2017-01-15
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    In the question above, $K$ is a separable $F$-algebra, and $E$ is a extension field of $F$. By a result in Proposition 7.4, $E\otimes _FK$ is a separable extension field of $E$. I know the tensor of two fields may be not a field... However, in question, the tensor $E\otimes _FK$ is a particular case... – 2017-01-15
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    What result are you using to conclude that $E\otimes_F K$ is a separable extension field? As I said before, in order to use Proposition 7.4 to conclude that something is a separable extension field, you need to already know that it is a _field_. – 2017-01-15
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    en.....you are right.... – 2017-01-15