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I had a so simple question.

Question: Let $E$ be an algebraic extension field of a field $F$. Does it follow that $E(x)$ is an algebraic extension field of $F(x)$?

I think it is not true and i can easily take counter example.

umm.. but i can't take one.

What's the example one?

Batominovski
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hew
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1 Answers1

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If $x$ is algebraic over $E$, then $x$ is also algebraic over $F$ since $E$ is an algebraic extension over $F$. Therefore $E(x)$ is algebraic over $F$. This shows that $E(x)$ is algebraic over $F(x)$ as well.

If $x$ is transcendent over $E$, then you should read up on transcendence degree. Write $\text{tr-deg}_K(L)$ for the transcendence degree of a field extension $L$ over a field $K$. Then, for a tower of fields $M\geq L\geq K$, we have $$\text{tr-deg}_K(M)=\text{tr-deg}_L(M)+\text{tr-deg}_K(L)$$ (see here). Thus, $$\text{tr-deg}_F\big(E(x)\big)=\text{tr-deg}_E\big(E(x)\big)+\text{tr-deg}_F(E)=1+0=1\,,$$ as $E$ is algebraic over $F$. On the other hand, $$\text{tr-deg}_F\big(E(x)\big)=\text{tr-deg}_{F(x)}\big(E(x)\big)+\text{tr-deg}_F\big(F(x)\big)=\text{tr-deg}_{F(x)}\big(E(x)\big)+1\,,$$ as $x$ is transcendent over $F$ as well. This shows that $$\text{tr-deg}_{F(x)}\big(E(x)\big)=0\,,$$ whence $E(x)$ is an algebraic extension of $F(x)$.

Batominovski
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