 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Take $a=[x]$. Every <quotient> class is a <polynomial> expression in $a$ with coefficients from $F$, so $K=F[a]$. Substitution gives $f(a)=[f(x)]=0$. Thus
$$
\boxed{K=F[a],\qquad f(a)=0.}
$$
This constructs a <field> containing a root even when no root existed in the original <field>.

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