Understanding morphism extensions theorem

Question

In Galois Theory class, we have seen the following result.

Let $K$ be a field and $F_1$, $F_2$ be two extension fields of $K$. Let $\alpha\in F_1$ be an algebraic element. Let $\beta \in F_2$. Then, the following two statements are equivalent:

1. $\exists f\colon K(\alpha)\to F_2$ s.t. $f|_K = \operatorname{id}$ and $f(\alpha)=\beta$;

2. $\beta$ is a root of $\operatorname{Irr}(\alpha,K)$ where $\operatorname{Irr}(\alpha,K)$ is the irreducible polynomial of $\alpha$ in $K$.

I am trying to understand this statement. If I chose $K=\mathbb{Q}$ with $F_1=\mathbb{Q}(\sqrt{2})$, $F_2=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and I define $$f\colon \mathbb{Q}(\sqrt{2})\to \mathbb{Q}(\sqrt{2},\sqrt{3}): \sqrt{2} \mapsto \sqrt{3},$$ then $\sqrt{3}$ should be a root of $x^2-2$ which is not.

• Am I wrong because my $f$ is not a morphism? In this case, how can I know that my $f$ fails to be a morphism?

• If the above point is not the case, what am I considering wrong?

Answer 1

Elements of $\mathbb Q (\sqrt 2,\sqrt 3) $ are of the form $a+b\sqrt2+c\sqrt3+d\sqrt6$ and elements of $\mathbb Q (\sqrt 2)$ are $a+b\sqrt2$, where $a,b,c,d\in\mathbb Q $. If you only demand $f (\sqrt 2)=\sqrt 3$, then $f(a+b\sqrt2)=a+c\sqrt 2+b\sqrt 3+d\sqrt 6$, where $c $ and $d $ are arbitrary rational numbers. Hence, your $f $ is not well defined.