Trivial question about splitting fields I'm having trouble with super basic ring theoretic concepts. Given some irreducible $f$ in $k[x]$, denote by $p$ the projection to the quotient $F$ by the ideal generated by $f$. I'm struggling to understand why $p(x)$ is now a root of $f$. The wiki page says $f(p(x))=p(f(x)) =0$, but I don't understand the left equality. What is going on? 
 A: It is as simple as it looks. We first have $p(0)=0$. Consider $\alpha\in k,\, \alpha\neq 0$. It is a polynomial of degree $0$. Perform the Euclidean division of $\alpha$ by $f$ to get $\alpha=0\cdot f+\alpha$. This means $\forall \alpha\in k,\,p(\alpha)=\alpha$ and therefore
$$\begin{align}p(f(x))&=p(\sum_0^n\alpha_i x^i)\\&=\sum_0^np(\alpha_i)p(x)^i\\&=\sum_0^n\alpha_ip(x)^i\\&=f(p(x))\end{align}$$
A: I also had a lot of trouble with this when I first saw it.  Here is the explanation that made sense to me with the details written out.  
If $k$ is a field, and $p(X)$ is an irreducible polynomial in the ring $k[X]$, then the ideal $\mathfrak p$ generated by $p(X)$ is a maximal ideal (because, for example, $k[X]$ is a principal ideal domain).  Thus the quotient ring $F = k[X]/\mathfrak p$ is a field.
This field contains $k$.  Well, not really.  But, the mapping $a \mapsto a + \mathfrak p$ gives an injective ring homomorphism $k \rightarrow F$, and so you can regard $k$ as a subfield of $F$, by identifying any $a \in k$ with the coset $a + \mathfrak p \in F$).  But to be precise, we will say that we have an injective ring homomorphism $\phi: k \rightarrow F$ given by $\phi(a) = a + \mathfrak p$.
The fields $k, F$ are respectively subsets of the polynomial rings $k[Y], F[Y]$.  Since we have a natural inclusion of $k$ inside $F$, we also have a natural inclusion of the polynomial ring $k[Y]$ inside the polynomial $F[Y]$ (I am using a different letter to denote a polynomial ring here, because $F$ itself is defined in terms of a polynomials in the indeterminate $X$).  That is, the injective homomorphism $\phi: k \rightarrow F$ extends to an injective homomorphism (let's also call it $\phi$) $k[Y] \rightarrow F[Y]$, given by $$\phi(a_0 + a_1Y + \cdots + a_nY^n) = \phi(a_0) + \phi(a_1)Y + \cdots + \phi(a_n)Y^n$$ $$ = (a_0 + \mathfrak p) + (a_1 + \mathfrak p)Y + \cdots + (a_n + \mathfrak p)Y^n$$ The image of this homomorphism is clearly $\phi(k)[Y]$, i.e. polynomials with coefficients in $\phi(k)$.  So basically, you can regard any polynomial with coefficients in $k$, as a polynomial with coefficients in $F$, in a natural way.   So now your polynomial $p(Y) \in k[Y]$ gives a polynomial $\phi(p(Y)) \in \phi(k)[Y] \subseteq F[Y]$.  
And the claim is that $\phi(p(Y))$ (a polynomial with coefficients in $\phi(k)$) has a root in $F$.  That root is the coset $X + \mathfrak p \in F$.  Write out $p(Y) = c_0 + c_1Y + \cdots + c_nY^n$ for $c_i \in k$, so that $$\phi(p(Y)) = (c_0 + \mathfrak p) + (c_1 + \mathfrak p)Y + \cdots + (c_n + \mathfrak p)Y^n$$ If you evaluate this polynomial at $X + \mathfrak p$, then from the fact that $fg + \mathfrak p = (f+ \mathfrak p)(g + \mathfrak p)$ and $(f+ \mathfrak p) + (g+ \mathfrak p) = (f + g) + \mathfrak p$ for any $f, g \in k[X]$, you get $$ (c_0 + \mathfrak p) + (c_1 + \mathfrak p)(X + \mathfrak p) + \cdots + (c_n + \mathfrak p)(X + \mathfrak p)^n $$ $$ = (c_0 + c_1X + \cdots + c_nX^n) + \mathfrak p = p(X) + \mathfrak p = 0 + \mathfrak p$$ because $p(X) \in \mathfrak p$.  
In textbooks and pretty much everywhere, they are not going to say "the claim is that $\phi(p(Y))$ has a root in $F$."  Instead, they will say "the claim is that $p(Y)$ has a root in $F$," because they are not going to bother distinguish $k$ from its isomorphic copy $\phi(k)$.  And in the same way they are going to think of $k[X]$ as a subring of $F[X]$, even though it is not really.  These identifications are natural and won't lead to any contradictions.  If you are comfortable with them, the proof becomes a lot shorter.
A: Suppose $f(x) = a_0 + a_1x + \cdots + a_nx^n \in F[x]$ for some field $F$.
If $u = x + \langle f(x)\rangle$ (the coset of $x$ in $F[x]/\langle f(x)\rangle$), then:
$f(u) = a_0 + a_1(x + \langle f(x)\rangle) + \cdots + a_n(x + \langle f(x)\rangle)^n$
...now a natural question is how we can multiply an element of $F[x]/\langle f(x)\rangle$ by an element of $F$?
What we do is define for $c \in F$: $c(g(x) + \langle f(x)\rangle) = (c\cdot g(x)) + \langle f(x) \rangle$.
This "makes sense" because the image $p(F)$ of $F$ under $p$ sends $c \mapsto c + \langle f(x)\rangle$ (the latter sees "$c$" as a CONSTANT polynomial) is an isomorphism of $F$ with $p(F)$.
So $f(u) = (a_0 + \langle f(x)\rangle) +  (a_1 + \langle f(x)\rangle)(x + \langle f(x)\rangle) + \cdots + (a_n + \langle f(x)\rangle)^n$.
Recall that in a quotient ring $R/I$, we have $(a+ I)(b + I) = ab + I$, so the above becomes:
$f(u) = (a_0 + \langle f(x)\rangle) + (a_x + \langle f(x)\rangle) +\cdots + (a_nx^n + \langle f(x)\rangle)$
And in a quotient ring $R/I$, we have $(a + I) + (b + I) = (a+b) + I$,
so we have:
$f(u) = (a_0 + a_1x + \cdots + a_nx^n) + \langle f(x)\rangle = f(x) + \langle f(x)\rangle$
But $f(x) + \langle f(x)\rangle = 0 + \langle f(x) \rangle = 0_{F[x]/\langle f(x)\rangle}$, since $f(x) = f(x) - 0 \in \langle f(x)\rangle$.
Thus $p(f(x)) = f(x) + \langle f(x)\rangle = 0_{F[x]/\langle f(x)\rangle} = f(u) = f(p(x))$.
A: $p(x)$ is confusing notation because it looks like you're applying a polynomial to $x$. I will use $\underline{a}$ instead for the image of $a \in k[x]$ in $k[x]/f(x)$. Also, I don't understand why everyone's working so hard. Write $f(x) = \sum f_n x^n$. The key point is that $a \mapsto \underline{a}$ is a $k$-algebra homomorphism, and this is true more or less by definition of the $k$-algebra structure on $k[x]/f(x)$. Now it follows that
$$\underline{f(x)} = \underline{\sum f_n x^n} = \sum f_n \underline{x}^n = f(\underline{x})$$
but since by definition $\underline{f(x)} = 0$ we conclude that $f(\underline{x}) = 0$ as well. 
