TITLE: If $L,K$ fields, and $L(\alpha)=K(\alpha)$ then $L=K$
QUESTION [7 upvotes]: The question appeared in my mind when I was studying Field Extension(preparation for Galois Theory).
It is trivial that if $L=K$, and $\alpha$ algebraic over $L$ and $K$, then $L(\alpha)$=$K(\alpha)$
Is it always true that If $L,K$ fields, $L(\alpha)=K(\alpha)$, and $[K(\alpha):K]=[L(\alpha):L]$  then $L=K$?
I think it is true and I try to prove it via the following statements:
It is clear that the smallest field containing $L$ and $\alpha$ is equal to the the smallest field containing $K$ and $\alpha$. Then the smallest field $F$ containing $L,K,\alpha$ is equal to both $L(\alpha)=K(\alpha)$.
Thus $F\subset L(\alpha)$ implies $K\subset L(\alpha)$. Then we can conclude that $K\subset L$
Similarly, $L\subset K$
Finally, $L=K$
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Thank you all guys for providing trivial or non-trivial examples, the statement is clearly false.

REPLY [8 votes]: This is False.
Consider the Field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ that, by Primitive Element Theorem, is generated by a single element $\alpha$ (You can choose $\alpha=\sqrt{2}+\sqrt{3}$, for example).
Now $\mathbb{Q}(\sqrt{2})(\alpha)$ is obviousely equal to $\mathbb{Q}(\sqrt{3})(\alpha)$ and $$[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{3})]=[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{2})]=2$$ but the fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are not isomorphic!