How do I show that $x$ is a root of $p(x)$ in $F[x]/\langle p(x)\rangle$? Let $F$ be a field, and $p(x)$ an irreducible polynomial in $F[X]$.
Define $K=F[X]/\langle p(x)\rangle$.
For every $a\in F$, denote $\bar{a}=\left<p(x)\right>+a$.
How do I show that $\bar{x}\in K$ is a root of $p(x)$?.
 A: You can think of $K=\Bbb Q[X]/\langle X^2-2\rangle$ as $\Bbb Q[\alpha]$ where $\alpha^2 = 2$. That is, $\alpha\in K$ is a root of the polynomial $f(X) = X^2-2$. Officially, $\alpha = \overline X\in K$, as $\overline X^2 - 2 = \overline{X^2-2} = \overline 0$. The same works for any irreducible polynomial (irreducibility is needed to prove that $K$ is a field).
A: Hint $\, $ The ring $\,\rm\color{#0a0}{hom}\,$ $\rm\:f\mapsto \bar f = f + \left<p\right> \:$  preserves $\rm\:\color{#0a0}{sums\,\ \&\,\ products}\:$ so it preserves $\rm\color{#c00}{polynomials}$ (which are compositions of sums and products). More explicitly  
$$ \begin{eqnarray}
\rm 0\, =\ \overline{\color{#c00}{p(x)}}\:
&=&\rm\ \  \overline{a_n x^n +\,\cdots + a_1 x + a_0}\\
&=&\rm\,\ \overline{a_n x^n}\, +\,\cdots + \overline{a_1 x} + \overline a_0\quad by\ \ \ \color{#0a0}{\overline{f+g}\ =\, \overline f + \overline g}\ \ \ \,\forall\ f,g \in F[x]\\
&=&\rm\,\  \overline a_n\,  \overline x^n+\,\cdots + \overline a_1\overline x + \overline a_0\quad by\ \ \ \color{#0a0}{\overline{f\, *\, g}\, =\, \overline f\, *\, \overline g}\ \ \ \forall\ f,g \in F[x]  \\
&=&\rm\  \bar{\color{#c00}p}(\overline{\color{#c00}x})\\
\end{eqnarray}$$
Remark $\ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $\,0$-ary operations), it follows by (structural) induction that homs also preserve the "polynomial" terms generated by the basic operations. Said equivalently, hom's commute with polynomials.
A: Let $\mathbb F$ be a field. Denote with $\mathbb F[x]$ its polynomial ring,
and let $p\in\mathbb F[x]$ be an irreducible polynomial. Observe the following:

*

*Because $\mathbb F$ is a field, $\mathbb F[x]$ is a PID (and is also a Euclidean domain).


*The irreducibility of $p$ implies that the principal ideal $\langle p\rangle\equiv p \mathbb F[x]$ is maximal. Quotient rings over maximal ideals are fields, thus $\mathbb F[x]/\langle p\rangle$ is a field.


*The field $\mathbb F$ (not the polynomial ring) is isomorphic to a subfield of $\mathbb F[x]/\langle p\rangle$ via the mapping
$$\varphi : \mathbb F \ni a \mapsto [a]\equiv a + \langle p\rangle \in \mathbb F[x]/\langle p\rangle,$$
where on the right-hand size, $a$ is to be understood as the constant polynomial $a\in\mathbb F[x]$. In other words, $\mathbb F$ is isomorphic to the subfield of $\mathbb F[x]/\langle p\rangle$ of constant polynomials modulo $\langle p\rangle$.
Identifying a polynomial with the sequence of its coefficients, then the $a$ in $[a]$ really stands for the sequence $(a,0,0,...)$.


*Therefore $\mathbb F[x]/\langle p\rangle$ is a field extension of $\mathbb F$, which means that we can "evaluate" polynomials $q\in\mathbb F[x]$ on equivalence classes $[s]\in\mathbb F[x]/\langle p\rangle$. In other words, expressions of the form $q([s])$ are meaningful.
The precise definition of this action is that, if $q(x)=\sum_k q_k x^k$, then
$$q([s]) = \sum_k [q_k] [s]^k = [\sum_k q_k s^k] \equiv [q(s)].$$
Note that in this expression $s$ is a polynomial, $s\in\mathbb F[x]$, so $q(s)$ here means to evaluate the polynomial $q\in\mathbb F[x]$ at the "point" $s\in\mathbb F[x]$ via the homomorphism $\mathbb F\to\mathbb F[x]$ (this is not technically a field extension because $\mathbb F[x]$ is not a field).


*Consider then the identity polynomial $x\equiv (0,1,0,...)\in\mathbb F[x]$, and the corresponding element $[x]\in\mathbb F[x]/\langle p\rangle$.
Then, as shown above, $p([x])=[p(x)]$. But
$$p(x) \equiv \sum_k p_k x^k = p.$$
Let me stress that in this expression we have, on the left-hand size, the polynomial $p$ acting on $x\in\mathbb F[x]$; this action gives as a result the polynomial $p$ itself, which is why the seemingly weird expression makes sense.
We conclude that $p([x])=[p(x)]=[p] = [0] \equiv \langle p\rangle$, which is what is meant when saying that "$x$ is a root of $p$ in $\mathbb F[x]/\langle p\rangle$".
