# Quadratic Extensions over Field of Characteristic $\neq 2$.

The following is an example taken from Dummit and Foote.

Let $F$ be a field of characteristic $\neq 2$ and let $K$ be an extension of $F$ of degree 2, $[ K : F]=2$. Let $\alpha \in K$ not contained in $F$. Then $\alpha$ satisfies an equation of degree at most $2$ over $F$. It follows that the minimal polynomial of $\alpha$ is the monic quadratic, $$m_{\alpha}(x)=x^2 + bx +c$$ Since $F \subset F(\alpha) \subseteq K$ and $K$ and $F(\alpha)$ have the same degree we have that $F(\alpha) = K$. The roots of the quadratic extension can be determined by the quadratic formula, $$\alpha = \frac{-b \pm \sqrt{b^2 - 4c} }{2}$$ Here $b^2 - 4c$ is not a square in $F$ because $\alpha$ in not an element of $F$.

Why does $\alpha \notin F$ imply the discriminant is not a square ?

Now $F(\alpha) = F \left ( \sqrt{b^2 - 4c} \right)$ as follows: by the formula above, $\alpha$ is an element of the field $F \left ( \sqrt{b^2 - 4c} \right)$

How does the quadratic formula give us this information? I'm assuming $F \left ( \sqrt{b^2 - 4c} \right)$ looks something like this, $b_0+b_1 \left ( \sqrt{b^2 - 4c} \right)$. Then $\alpha$ is in this field extension if we know that $\pm \frac{b}{2} \in F$. How do we know this is true?

• $b\in F$ so $\frac{b}{2}$ lies in $F$ too. – Daniel Jul 20 '15 at 0:42

If $b^2-4c$ were a square in $F$, say $k^2$ where $k\in F$, then $$\alpha=\frac{-b\pm k}2$$ would be also in $F$, because $\alpha$ can be calculated with internal operations over $F$. This contradicts the previous assumption, so $b^2-4c$ is not a square.
Now, since $\alpha$ is in $K$, we have $$[K:F(\alpha)]\cdot[F(\alpha):F]=[K:F]=2\qquad(*)$$
Now, $\alpha$ is a root of a quadratic polynomial over $F$, so $[F(\alpha):F]$ is $1$ or $2$, but $\alpha\notin F$, so $[F(\alpha):F]=2$. This, with $(*)$, implies that $[K:F(\alpha)]=1$, thus $K=F(\alpha)$.
Let $\beta=\sqrt{b^2-4c}$. Note that $\alpha=\frac{-b\pm\beta}2$ and $\pm\beta=2\alpha+b$, so $F(\alpha)=F(\beta)=K$.
• Overall, I am not sure what dividing by $2$ in an arbitrary field means. Does $2$ even need be an element of $F$ ? – user7090 Jul 20 '15 at 0:52
• @asdvn Every field has a multiplicative identity 1, and you can define $2=1+1$ for an arbitrary field. This is why you need the characteristic to be different from 2, because in this example you want to be able to divide by 2, and 2 has an inverse in a field of and only if $2\neq 0$. – coldnumber Jul 20 '15 at 1:21