# Counterexample: If $f: \mathbb{Q}(\alpha) \to \mathbb{Q}(\alpha)$ is an isomorphism of fields, then $\beta=f(\alpha)$. [closed]

State whether the statement below is true or false:

If $f: \mathbb{Q}(\alpha) \to \mathbb{Q}(\beta)$ is an isomorphism of fields, then $\beta=f(\alpha)$.

If true, provide a proof; if false, provide a counter-example.

I know that this statement is false but I can't think of a counter-example.

## closed as off-topic by Thomas Andrews, Andrés E. Caicedo, 6005, Mark Viola, Harish Chandra RajpootNov 5 '15 at 5:06

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• What's $\beta$? Did you mean to speak of an isomorphism $f:\mathbb Q(\alpha) \rightarrow \mathbb Q(\beta)$ – lulu Nov 5 '15 at 2:24
• What is $\beta$? Is $\alpha$ algebraic? – Thomas Andrews Nov 5 '15 at 2:24
• $\beta$ and $\alpha$ are not specified in this problem, and $f$ is an isomorphism – Username Unknown Nov 5 '15 at 2:26

• Ok to make sure. So when I proving that this is indeed a counterexample, I would do the following $$f(\alpha-1)=f(\alpha)-f(1)=\beta-1\neq \beta$$ – Username Unknown Nov 5 '15 at 2:41
For an example where $Q(a)\ne Q$, let $a=\sqrt 2$ and $b= -a$. We have $Q(a)=Q(b)=\{x+y a :z,y\in Q\}$. Every $z\in Q(a)$ is equal to $x_z+y_z a$ where $x_z , y_z \in Q$ and are uniquely determined by $z$. Let $f(x_z+y_za)=x_z-y_za$. This is a field isomorphism (a certain kind,called an involution, because $f(f(z))=z)$.Observe that $f(a)=b\ne a$.