# Field extension question, basis, cannot be expressed as linear combination?

Suppose I have a field extension K of F with basis $\{1,\beta\}$, $\beta\in K^*/F^*$.

How do I show that $\beta^2$ cannot be written as $c_1+c_2\beta$, where $c_1,c_2\in F, c_2\ne 0$ unless $\beta^2 \in F^*$?

For example, if $K=\mathbb{Q}(\sqrt 2)$, $F=\mathbb{Q}$, I would want to show that $2$ cannot be written as $c_1 +c_2 \sqrt{2}$, where $c_i \in \mathbb{Q}, c_2\ne 0$, unless 2 is in $\mathbb{Q^*}$, which it is. This follows by arguing that $\sqrt{2}$ is irrational, but how do I do it in the more general setting as in above?

Sincere thanks for help.

-
But 2 is $2+0\sqrt{2}$. Take $c_1=2,c_2=0$. – Sigur Sep 2 '12 at 15:27
If $\{1,\beta\}$ is a basis for $K$ over $F$ then every element of $K$ can be written as $c_1+c_2\beta$ where $c_1,c_2 \in F$ in particular $\beta^n$. – JSchlather Sep 2 '12 at 15:29
@Sigur That was fast! I edited my question above. – yoyostein Sep 2 '12 at 15:29
@JacobSchlather Thanks (+1), but I just added a new condition that $c_2\ne 0$. – yoyostein Sep 2 '12 at 15:32
The intended question seems to be muddled. If $\{1, \beta\}$ are an $F$-basis of $K$, then necessarily $\beta^2$ can be written (uniquely) as $c_1+c_2\beta$ with $c_1, c_2\in F$. Then $X^2-c_2 X - c_1$ is the minimal poylnomial of $\beta$. As it is irreducible, we conclude that is $c_1=0$ is impossible. However, $c_2=0$ is equivalent to $\beta^2=c_1\in F$. – Hagen von Eitzen Sep 2 '12 at 17:14

This is still false. Let $F=\mathbb{Q}$, $K=F(\sqrt{2})$ and $\beta=1+\sqrt2$. Then $$\beta^2=1+2\sqrt2+2=3+2\sqrt2=1+2\beta.$$ Here $c_2=2\neq0$ as requested, but $\beta^2\notin F$.