Question

If $\mathbf{p}\left(\mathbf{x}\right)$is an irreducible quadratic polynomial in $\mathbf{F}\left[\mathbf{x}\right]$, show that$\raisebox{1ex}{$\mathbf{F}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$ contains all the roots of$\mathbf{p}\left(\mathbf{x}\right)$.

Short answer

It is proved that$\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$ contains all the roots of $p\left(x\right)$.

Step-by-step solution

Statement of theorem 5.11

Theorem 5.11states consider that$F$ as a field,$p\left(x\right)$ as an irreducible polynomial in $F\left[x\right]$. Then$\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$ willbe an extension field$K$ of role="math" localid="1649763074612" $F$, which contains a root of $f\left(x\right)$.

Show that F[x](px) contains all the roots of

Theorem 5.10states consider that $F$as a field and $p\left(x\right)$ a nonconstant polynomial in $F\left[x\right]$. Then the statements are equivalent as follows:

(1) $p\left(x\right)$is irreducible in $F\left[x\right]$.

(2) $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$will be a field.

(3) $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$will be an integral domain.

According to theorem 5.11, there is $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$, which has the root $\left[x\right]$of $p\left(y\right)$. As a result, $y-\left.\left[x\right]\right|p\left(y\right)$because $p\left(y\right)$will be quadratic its quotient by $y-\left[x\right]$be linear.

Therefore,$\frac{p\left(y\right)}{y-\left|x\right|}=\left[a_{1}x+a_0\right]y+\left[b_{1}x+b_0\right]$ with $\left[a_{1}x+a_0\right]\ne \left[0\right]$.

Now, according to theorem 5.10, $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$will be a field, and the other root of $p\left(x\right)$is $\frac{\left[{\mathrm{b}}_1\mathrm{x}+{\mathrm{b}}_0\right]}{\left[a_{1}x+a_0\right]}\in \raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$. All of these are roots of $p\left(x\right)$.

Hence, it is proved that$\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(p\left(x\right)\right)$}\right.$contains all the roots of $p\left(x\right)$.