Help interpreting a question about irreducible polynomials I am asked the following:
If $p$ is a monic irreducible quadratic in $F[x]$, where $F$ is a field, show that $p$ factors into linear factors over $E=F[x]/\langle p\rangle$
However I am having trouble interpreting the second part of the statement.
To me it sounds like we take $p\in F[x]$ and then evaluate it on the field $E$, that is to say, for the second time $p$ is mentioned the coefficients are members of $E$, thus the second time $p \in E[x]$.
Due to how convoluted this sounds I suspect my interpretation of the question to be wrong. What is the correct way of reading this?
 A: First of all: Yes, your interpretation is correct. To say that $p$ factors over $E$ is a common way to express the fact that when you interpret $p$ as a polynomial with coefficients in $E$, i.e. as an element of $E[x]$, then it does factor.
This can be tricky to wrap your head around in the abstract setting. But let's consider $p=x^2+1\in\Bbb R$, so $F=\Bbb R$ and then we get $E=\Bbb R[x]/\langle x^2+1\rangle=\Bbb C$. I am hoping that the last equality there is clear to you already, essentially the residue class of $x$ in $\Bbb R[x]/\langle x^2+1\rangle$ becomes $i$ because it satisfies $i^2+1=0$. Now, clearly the polynomial $p$, considered as a polynomial with complex coefficients, does factor over $\Bbb C$ as $p=(x+i)(x-i)$.
This might be much less confusing because you can think in terms of $i$ instead of the residue class of $x$ in $\Bbb R[x]/\langle x^2+1\rangle$ - however, once you walk through each of the steps in this particular example, it should be easier to generalize.
Let's write $E=F[\xi]$, where $\xi$ is the residue class of $x$ in $F[x]/\langle p\rangle$. If $p=x^2+ax+b$, then this means $\xi^2+a\xi+b=0$. In other words, you may say, $p(\xi)=0$, so $p$ has a zero at $\xi$. 
A: We have
$p(x) \in F[x]; \tag 1$
that is, the coefficients of
$p(x) = \displaystyle \sum_0^n p_i x^i \tag 2$
all lie in $F$:
$p_i \in F, \; 0 \le i \le n. \tag 3$
We are also given that $p(x)$ is irreducible in $F[x]$; in particular, $p(x)$ has no zeroes in $F$; for all $\alpha \in F$, $p(\alpha) \ne 0$.
Now suppose $G \supset F$ is any extension field of $F$; since $p(x) \in F[x]$, we also have $p(x) \in G[x]$, but in general $p(x)$ will no longer be irreducible when considered as an element of $G[x]$, since the "extra coeffiecients" available in $G$ may allow $p(x)$ to be factored in $G[x]$; a good example is provided by the polynomial
$r(x) = x^4 + 1 \in \Bbb Q[x]; \tag 4$
$r(x)$ is irreducible over $\Bbb Q$, that is, in $\Bbb Q[x]$, by an application of Eisenstein's Criterion with $p = 2$; though we can't apply Eisenstein directly, if we consider the polynomial
$r(x + 1) = (x + 1)^4 + 1 = x^4 + 4x^3 + 6x^2 + 4x + 1 + 1 = x^4 + 4x^3 + 6x^2 + 4x + 2, \tag 5$
we can apply the criterion with $p = 2$ to conclude that $r(x + 1)$, hence $r(x)$, is irreducible in $\Bbb Q[x]$.  However, if we adjoin $\sqrt 2$ to $\Bbb Q$, we find that in $\Bbb Q(\sqrt 2)[x]$ we have
$r(x) = x^4 + 1 = (x^2 + \sqrt 2 x + 1)(x^2 - \sqrt 2 x + 1); \tag 6$
another, perhaps even more germane example is provided by
$c(x) = x^3 - 2 \in \Bbb Q[x]; \tag 7$
a direct application of Eisenstein with $p = 2$ indicates that $c(x)$ is in fact irreducible in $\Bbb Q[x]$; however, if we adjoin $\sqrt[3] 2$ to $\Bbb Q$, we find that in $\Bbb Q(\sqrt[3] 2)[x]$,
$c(x) = x^3 -2 = (x - \sqrt[3] 2)(x^2 + \sqrt[3] 2 x + (\sqrt[3] 2)^2); \tag 8$
$c(x)$ cannot be reduced further in $\Bbb Q(\sqrt[3] 2)[x]$ because any factor of degree less than $2$ must be linear, and the roots of $x^2 + \sqrt[3] 2 x + (\sqrt[3] 2)^2$ are complex, and $\Bbb Q(\sqrt[3] 2)$ is a real field.
The upshot of these remarks--what is germane to the problem at hand and to our OP Makogan's quest to properly interpret certain statements in the body of question--is that though the coefficients of polynomials such as $p(x)$, $r(x)$ or $c(x)$ may lie in a base field such as $F$ or $\Bbb Q$, the arithmetic operations implicitly prescribed in an expression such as (2), (6) or (8), may be performed in appropriate extension fields, and may yet produce resulting coefficients which lie in the base.  This notion is, in a sense, the essence of the idea that we may factor irreducible polynomials via extending their fields of coefficients, and of course this may always be done in abstracto by forming a field of the general form $F[x] / \langle q(x) \rangle$, where $q(x) \in F[x]$ is irreducible. 
Applying these remarks to the present problem, with the monic quadratic
$p(x) = x^2 + p_1 x + p_0 \tag 9$
irreducible in $F[x]$, it is also prime and hence the ideal $\langle p(x) \rangle \subset F[x]$ is prime, hence maximal since $F[x]$ is a principal ideal domain; thus $F[x] / \langle p(x) \rangle$ is a field.  If
$t = x + \langle p(x) \rangle \in F[x] / \langle p(x) \rangle, \tag{10}$
is the $x$-coset of $\langle p(x) \rangle$, then
$t^2 + p_1 t + p_0 = (x^2 + \langle p(x) \rangle) + (p_1 x + \langle p(x) \rangle) + (p_0 + \langle p(x) \rangle)$
$= (x^2 + p_1 x + p_0) + \langle p(x) \rangle = \langle p(x) \rangle = 0 \in F[x] / \langle p(x) \rangle, \tag{11}$
so $t$ is a root of $p(x)$ in the field $F[x] / \langle p(x) \rangle$; therefore $x - t \mid x^2 + p_1 x + p_0$ over this field, and evidently the quotient is a monic polynomial of degree $1$; therefore there exists some $u \in F[x] / \langle p(x) \rangle$ with
$(x - t)(x - u) = p(x) = x^2 + p_1 x + p_0; \tag{12}$
this yields, upon comparing coefficients,
$-t - u = p_1, \; tu = p_0, \tag{13}$
or
$t + u = -p_1, \; tu = p_0; \tag{14}$
having $t$, we thus write $u$ as
$u = -t - p_1, \; \text{or} \; u = p_0 t^{-1}; \tag{15}$
then we see that 
$(x - t)(x + (t + p_1)) = x^2 + (p_1 + t - t) x - t(t + p_1) = x^2 + p_1 x + p_0 = p(x), \tag{16}$
since 
$- t(t + p_1) = -t^2 - p_1 t = p_0 \tag{17}$
from (11).
We thus see that $p(x)$ splits as in (12) or (16) over the field $F[x] / \langle p(x) \rangle$.
