Exercise 2.4.6

Question

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This exercise will describe how to solve quadratic equations over a field $F$ of characteristic 2.
\begin{enumerate}
\item[(a)] Given $b \in F$, we will assume there is a larger field $F\subset L$ such that $b=\beta^2$ for some $\beta \in L$.
Show that $\beta$ is unique and that $\beta$ is the unique root of $x^2+b$. Because of this, we denote $\beta$ by $\sqrt{b}$.
\item[(b)] Now suppose that $f = x^2+ax+b$ is a quadratic polynomial in $F[x]$ with $a
e 0$. Suppose also that $f$ is irreducible over $F$, so that it has no roots in $F$. We will see in Chapter 3 that $f$ has a root $\alpha$ in a field $L$ containing $F$. Prove that $\alpha$ cannot be written in the form $\alpha = u + v \sqrt{w}$, where $u,v,w \in F$.
\item[(c)] Part (b) shows that solving a quadratic equation with nonzero $x$-coefficient requires more than square roots. We do this as follows. If $b\in F$, let $R(b)$ denote a root of $x^2+x+b$ (possibly lying in some larger field). We call $R(b)$ and $R(b)+1$ the $2$-roots of $b$. Prove that the roots of $x^2+x+b$ are $R(b)$ and $R(b)+1$, and explain why adding 1 to the second $2$-root gives the first. 
\item[(d)] Show that the roots of $f = x^2 + ax+b, a
e 0$, are $aR(b/a^2)$ and $a(R(b/a^2)+1)$.
\end{enumerate}

Richard Ganaye · Accepted Answer

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\begin{proof}
\begin{enumerate}
\item[(a)]
Let $L$ an extension of $F$ and $\beta \in L$ such that $\beta^2 = b$.

As $x^2 - b = x^2 - \beta^2 = (x - \beta)^2$, $\beta$ is the unique root of $x^2-b = x^2+b$. We write $\beta = \sqrt{b} \in L$.

\item[(b)]
Suppose that  $f = x^2+ax+b, \ a
e 0$ is irreducible on $F$. As $\deg(f) = 2$, this is equivalent to the fact that $f$ has no root in $F$.
 $f $ has a root $\alpha$ in an extension $L \supset F$.
 
If $\alpha = u+v\sqrt{w},\ u,v,w \in F$, then $v
eq 0$, otherwise $\alpha \in F$, in contradiction with the irreducibility of $f$.
 
 Then
 \begin{align*}
 0&= \alpha^2 + a \alpha + b\
 &= u^2+wv^2 + a(u+v \sqrt{w})+ b\
 &= u^2+wv^2+au+b + av \sqrt{w}\
 &=s + t \sqrt{w},
 \end{align*}
 
where $s = u^2+wv^2+au+b  \in F, t = av \in F, t
eq 0$.

Thus $\sqrt{w} = -s/t \in F$, so $\alpha \in F$, in contradiction with the irreducibility of $f$.

Conclusion : $\alpha = u+v\sqrt{w},\ u,v,w \in F$ is impossible.

\item[(c)]
Write $R(b)$ a root of $x^2+x+b$ in an extension of $F$.

As $R(b)^2+R(b)+b=0$, $(R(b)+1)^2 + (R(b) +1)+b = R(b)^2+1 + R(b)+1+b = R(b)^2+R(b)+b=0$.

As $R(b)+1+1 = R(b)$, the two (distinct) roots of  $x^2+x+b$ are $R(b),R(b)+1$, and  $\sigma : x\mapsto x+1$ exchanges the two roots.

\item[(d)]

For all $y \in L$,
\begin{align*}
f(y) =0 &\iff y^2+ay+b=0\
&\iff \left(\frac{y}{a}ight ) ^2 + \left(\frac{y}{a}ight ) + \frac{b}{a^2} = 0\
&\iff \frac{y}{a} \in \left \{R \left(\frac{b}{a^2}ight ) , R \left(\frac{b}{a^2}ight ) +1 ight \}\
&\iff y \in \left \{aR \left(\frac{b}{a^2}ight ) , a\left [R \left(\frac{b}{a^2}ight ) +1ight] ight \}.\
\end{align*}
The roots of $x^2+ax+b,a
eq 0$ are so $aR \left(\frac{b}{a^2}ight ) , a\left [R \left(\frac{b}{a^2}ight ) +1ight] $.
\end{enumerate}
\end{proof}

Exercise 2.4.6

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Exercise 2.4.6

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