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If $K$ and $L$ are two extensions of a field $F$ then they are said to be $F$-isomorphic if there exists an isomorphism $\phi : K\rightarrow L$ which when restricted to the subfield $F$ forms an identity map.

Similarly I assume that $F$-homomorphism is a similar type of homomorphism between two field extensions which when restricted to the subfield $F$ forms an identity map.

I have two statements about field homomorphisms both of them false. But I don't know how they are false

$a)$ Let $K$ and $L$ be two field extensions of the field of rational numbers $Q$ then there exits a field homomorphism $\phi :K \rightarrow L$ which is not a $Q$-homomorphism

$b)$ Let $K$ and $L$ be two field extensions of the field $F=Q(\sqrt{2})$ then any field homomorphism $\phi :K \rightarrow L$ is a $F$-homomorphism

Can anyone please show me how these statements are false ?

Siddharth Prakash
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    The action of $\phi$ on $\Bbb{Q}$ is determined by $\phi(1)$. The action on $\Bbb{Q}(\sqrt2)$ is determined by $\phi(1)$ and $\phi(\sqrt2)$. – reuns Mar 18 '21 at 16:56

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A somewhat expansion of @reuns’s comment:

Your a) is wrong because if $K$ is a quadratic extension and $L$ is a cubic extension of $\Bbb Q$, then there are no field homomorphisms from $K$ to $L$.

Your b) is wrong because, for a minimal example, you can take $K=L=F$ and the field homomorphism $\varphi(a+b\sqrt2\,)=a-b\sqrt2$, for $a,b\in\Bbb Q$.

Lubin
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