Example of Basic Theorem of field extensions I'm studying the proof of Basic Theorem of field extensions from Pinter's A Book Of Abstract Algebra.
Theorem: Let $F$ be a field and $a(x)$ a nonconstant polynomial in $F[x]$.
There exists an extension field $E$ of $F$ and an element $c$ in $E$ such that $c$ is a root of $a(x)$.
Proof: To begin with, $a(x)$ can be factored into irreducible polynomials in $F[x]$. If $p(x)$ is any non-constant irreducible factor of $a(x)$, it is clearly sufficient to find an extension of $F$ containing a root of $p(x)$, since such a root will also be a root of $a(x)$.
If $p(x)$ is irreducible in $F[x]$, then $〈p(x)〉$ is a maximal ideal of $F[x]$. Furthermore, if $〈p(x)〉$ is a maximal ideal of $F[x]$, then the quotient ring $F[x]/〈p(x)〉$ is a field.
It remains only to prove that $F[x]/〈p(x)〉$ is the desired field extension of $F$. When we write $J = 〈p(x)〉$, let us remember that every element of $F[x]/J$ is a coset of $J$. We will prove that $F[x]/J$ is an extension of $F$ by identifying each element a in $F$ with its coset $J + a$.
To be precise, define $h: F → F[x]/J$ by $h(a) = J + a$. Note that $h$ is the function which matches every $a$ in $F$ with its coset $J + a$ in $F[x]/J$. We will now show that $h$ is an isomorphism.
I'm trying to have a better understanding of the proof and the concept. Wonder if someone could give me a SIMPLE specific example of the following so I have better grasp of the proof:
1) $a(x)$
2) $p(x)$
3) $c$
4) $F[x]/〈p(x)〉$
5) $〈p(x)〉 + a$
Thank you!
 A: The "classic" example is the following (I could devise an example with $a(x) \neq p(x)$, but we'll just deal with $p(x)$ for now):
Let $F = \Bbb R$, the field of real numbers. Let $p(x) = x^2 + 1$. It follows from the order properties of $\Bbb R$, that for any $r \in \Bbb R$, that $r^2 \geq 0$.
This polynomial ($x^2 + 1$) has no root in $\Bbb R$, and since it is a quadratic, is therefore irreducible (since any factor of lesser degree would have to be of the form: $t(x - \alpha)$, where $\alpha$ is a root).
So we consider the quotient ring $\Bbb R[x]/J$, where $J$ is the (principal) ideal generated by the polynomial $x^2 + 1$. A general element of this quotient can be put in the form:
$a + bx + J$, for example:
$x^2 + x + 1 + J = x + (x^2 + 1) + J = x + J$, since $(x^2 + x + 1) - x \in J$
(Here, we have $a = 0$, and $b = 1$).
Note that for $r \in \Bbb R$, the mapping $r \mapsto r + J$ is a ring-homomorphism (this just captures the well-known fact, that the constant term of the sum of two real polynomials is the sum of the constant terms, and that the constant term of the product of two real polynomials is the product of their constant terms).
In fact, this is an embedding of $\Bbb R$ into $\Bbb R[x]/J$, since the only real number that maps to $0 + J$ (which can also be thought of as $0 + 0x + J$) is the real number $0$. So we perform this kind of "mental translation" where we think of $r + J$ as being the real number $r$ (This is very much like thinking of the rational number $\dfrac{2}{1}$ as being the integer $2$, even though integers and rational numbers are two different beasties). Let's call this new field (isomorphic as a ring, and thus as a field, to $\Bbb R$) something like $R'$.
So, really, we're going to find a root of a polynomial in the ring $R'[x]$ (which is isomorphic to $\Bbb R[x]$). What happens next feels like sleight-of-hand to some people:
Consider the coset of $x$: that is, $x + J$. We compute $(x + J)^2 + (1 + J)$:
$(x + J)^2 + (1 + J) = (x + J)(x + J) + (1 + J) = (x^2 + J) + (1 + J) = (x^2 + 1) + J = 0 + J$
So $x + J$ is an element of $\Bbb R[x]/J$ that squares to $-1 + J$.
It is common practice to denote the coset $(a + bx) + J$ abbreviated as $a + bi$, and we identify $\Bbb R[x]/\langle x^2 + 1\rangle$ with the complex numbers, $\Bbb C$.
Verify for yourself, that we have:
$[(a + bx) + J][(c + dx) + J] = [(ac - bd) + (ad + bc)x] + J$, corresponding to the usual multiplication of complex numbers:
$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$
A: Take ur field to be $Q$ and ur $\alpha(X)$ to be $X²-2$. Then we have $Q[X]/(X²-2)$ has a root of $X²-2$ namely $\bar{X}$, the image of $X$ under the canonical homomorphism. Observe that  $Q[X]/(X²-2)$  is isomorphic to $Q[√2]$
