Field extension that contains root For a field $F$ and irreducible polynomial $p(x) \in F[x]$, the field extension $K = F[x]/(p(x))$ is supposed to have a root of $p(x)$. It is supposed to be the element $x + (p(x))$ in $K$, and the reason is supposed to be that $$p(x + (p(x))) = p(x) + (p(x)) = 0.$$ I understand why the second equality is true, but I do not see the first equality. I understand that there is a natural homomorphism from $F[x]$ to $K$, but I do not see how this implies the equality. Could someone explain this?
 A: Write $\bar x = x + I$, where $I=\langle p(x)\rangle$.
Then $p(\bar x) = p(x+I) = p(x)+I =I$, since $p(x)\in I$. Thus $p(\bar x) = \bar 0$ in $F[x]/I$.
Claim that $p(x+I) = p(x)+I$.
Indeed, write $p(x) = a_nx^n+\ldots+a_1x+a_0$.
Note that the operations on the quotient ring are defined as
$$(f+I) + (g+I) = (f+g)+ I,\quad (f+I)*(g+I) = (fg)+I.$$
Then $$p(x+I) = a_n(x+I)^n + \ldots a_1(x+I) + a_0 = (a_nx^n +\ldots + a_1x+a_0) + I = p(x)+I$$
since $(x+I)*(x+I) = x^2+I$ and so $(x+I)^k = x^k+I$ for each $k\geq 0$ and thus $a_k(x+I)^k = a_k(x^k+I) = a_kx^k + I$
A: You know about
$$
a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1}).
$$
This directly implies
$$
p(a)-p(b)=(a-b)q(a,b)
$$
for some polynomial $q$. This also gives if $b=a+m(x)p(x)$ in some extension of $K[x]$ that
$$
p(a+m(x)p(x))-p(a)=m(x)p(x)q(a+m(x)p(x),a)\in(p(x))
$$
In other words,
$$
p(a+(p(x)))\subset p(a)+(p(x)).
$$
Now set $a=x$ to get the (corrected) claim
$$
p(x+(p(x)))\subset p(x)+(p(x))=0+(p(x)).
$$
A: Another way to see this is just to convince yourself that the quotient map is a ring homomorphism. That is, $\phi: F[X] \rightarrow K = F[x]/(p(X))$ is a ring homomorphism. Once, you have this, your question becomes, why $\phi(p(X)) = p(\phi(X))$ as we want to essentially see $\bar{p(X)} = p(\bar{X})$ as the quotient map maps each element to the class of its element in the quotient. So if we were to write $p(X) = a_nX^n + \ldots a_1X + a_0$ then $\phi(p(X)) = \phi(a_nX^n + \ldots + a_1X + a_0) = \phi(a_nX^n) + \ldots + \phi(a_1X) + \phi(a_0) = a_n\phi(X)^n + \ldots + \phi(a_0).$ Explanations above essentially show why the natural map is a homomorphism, so it addresses your concern of seeing the implication just from the natural map. 
