Exercise 3.3.5 (Quadratic reducibility)

Question

If I gave you a quadratic over $\mathbb{Q}$, how would you decide
whether it was reducible or irreducible?

Hassaan Naeem · Accepted Answer

extbf{Solution:}
By 	extbf{Lemma 3.3.1 (ii)}, if the quadratic has a root in $\mathbb{Q}$, then it is reducible.
By the same lemma 	extbf{(iii)}, if the quadratic has no root in $\mathbb{Q}$, then it is irreducible.

Richard Ganaye · Answer

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} By Lemma 3.3.1, a polynomial $f(x) = a x^2 + bx + c \in \Q[x]$, where $a 
e 0$, is irreducible if and only if $f$ has no root in $\Q$.

Now $f(x) = a\left[ \left( x+ \frac{b}{2a}ight)^2 - \frac{\Delta}{4a^2} ight]$, where $\Delta  = b^2 - 4ac$.

If $f$ has a root $\alpha \in \Q$, then $\Delta = (2 \alpha + b)^2$ is a square in $\Q$. Conversely, if $\Delta$ is a square in $\Q$, $\Delta = \delta^2$, with $\delta \in \Q$, and the roots of $f$ are $\frac{-b \pm \delta}{2a}$, both in $\Q$.

We have proved that
$$ax^2 + bx + c 	ext{ is irreducible in } \Q[x] \iff \Delta = b^2 - 4ac 	ext{ is not a square in }\Q.$$

Note: $\Delta = u/v,\ u\in \Z, v \in \N^*$ is a square in $\Q$ if and only if $u,v$ are squares in $\N$.
\end{proof}

Exercise 3.3.5 (Quadratic reducibility)

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Exercise 3.3.5 (Quadratic reducibility)

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