Constructing a field in which polynomial has root 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.

 A: *

*The natural map $\iota\colon F\longrightarrow F[x]/(g)$ defined by $\iota(a)=a+(g)$ is injective. So, $\iota(F)$ is isomorphic to $F$ and therefore, although $F$ is not actually a subset of $F[x]/(g)$, it is natural te see $F[x]/(g)$ as an extension of $F$.

*Because $g(\alpha)=g\bigl(x+(g)\bigr)=g(x)+(g)=(g)$, and $(g)$ is the $0$ element of $F[x]/(g)$.

A: Consider the sequence of canonical morphisms $F\to F[X] \to F[X]/(g) = F_1$. These are ring morphisms so the composite is as well : but a ring morphism between fields is precisely a field morphism; those are known to be injective: hence we actually have an embedding $F\to F_1$ which allows us to see $F_1$ as an extension of $F$.Call this embedding $\iota$, and the embedding $F\to F[X]$ is $i$: $\iota = \pi\circ i$
Now if $g= \displaystyle\sum_{k=0}^d a_kX^k$, then $\iota(g)(\alpha)= \displaystyle\sum_{k=0}^d\iota(a_k)\alpha^k = \displaystyle\sum_{k=0}^d\pi(i(a_k))\pi(X)^k = \pi(\displaystyle\sum_{k=0}^di(a_k)X^k)= \pi(g) = 0$. Hence $\alpha$ is a root of $\iota(g)$.
But if we identify $F$ and $\iota(F)$ through $\iota$ this precisely means that $\alpha$ is a root of $g$.
There's no magic happening here: we want to find a root of $g$ so we add an element to $F$ (we get $F[X]$) and declare :"this new element is a root of $g$" (we kill $(g)$ by modding it out)
A: 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.
A: Why is $F_1$ a field extension of $F$
Properly speaking, $F_1$ is not an extension of $F$. However, there is a trivial embedding $f: t \mapsto p(x) =t$ which maps an element of $F$ into the polynomial of degree $0$ with constant term equal to $t$. You can then use the quotient map to embed $f(F)$ in $F_1$.
Why is $\alpha$ a root of $g$?
Because the embedding above maps $\alpha$ to the class of $x$ and $g(x)$ is the zero of $F_1$.
A: Firstly, $F$ is isomorphic to $\bar{F}=F+(g)\subset F_1$.
Secondly, and briefly, $g(x)=0\implies x$ is a root of $g$ (in an abstract,  algebraic sense;  just as $\mathbb R[x]/(x^2+1)$ is an algebraic version of $\mathbb C$).
Lastly remember that in $F_1$ the role of $x$ is played by $\bar{x}=x+(g)$...  its image under the quotient map... 
