TITLE: Constructing a field in which polynomial has root
QUESTION [6 upvotes]: The problem
Say we have a field $F$ and an irreducible polynomial $g \in F[x]$ of degree $\geq 1$. Let $(g)$ denote the (maximal) ideal generated by the $g$ in $F[x]$. Then define the field extension $F_1 = F[x]/(g)$ of $F$. Then $g$ has a root $\alpha$ in $F_1$, being $x + (g)$.
My question(s)

Why is $F_1$ a field extension of $F$? I don't see how the field $F$ can be contained in a quotient ring. I do know that $F_1$ is a field because $(g)$ is a maximal ideal in $F[x]$ so no need to explain this.
Why is $\alpha$ a root of $g$? Perhaps because $\pi(g(x)) = g(\pi(x))$ with $\pi$ the surjection from $F[x]$ to $F_1$? This would only work in my eyes if $\pi$ 'fixes' constants but it doesn't to that or I don't understand why it should.

REPLY [5 votes]: There is a low budget way to do this. You may be familiar with taking the Gaussian integers as 2 by 2 matrices; this is the same procedure, but allowing rational coefficients. Now that I think of it, the same procedure with real number coefficients gives the complex numbers.
Suppose we have an irreducible polynomial over the rationals such as
$$  g(x) = x^3 + x^2 - 2 x - 1.  $$
We construct the "companion matrix"
$$
A =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 2 & -1
\end{array}
\right)
$$
The matrix $A$ will stand as the representative of your root $\alpha.$
The elements of the field are, given rational coefficients $u,v,w,$ just
$$ uI + v A + w A^2 =
\left(
\begin{array}{rrr}
u & v & w \\
w & u + 2 w & v - w \\
v - w & 2v - w & u - v + 3w
\end{array}
\right)
$$
The rational numbers are into this matrix ring as $rI.$ The multiplicative identity is $I.$ Addition is obviously closed. Multiplication uses the identities
$$ A^3 = I + 2 A - A^2  $$
$$ A^4 = -I - A + 3 A^2  $$
There are no zero divisors since $x^3 + x^2 - 2 x - 1$ was irreducible. Multiplication is commutative because everything is a power of $A.$
The ring is a field since can find the reciprocal of anything nonzero, for instance
$$ A^{-1} = A^2 + A - 2 I.   $$ 
Finally $\alpha$ is a root because of Cayley-Hamilton.