I'm reading a theorem from my lectures, which is a sort of a preparatory theorem for the construction of the splitting field. I came across a part that I don't understand:
Let $K$ be a field and let $f \in K[X]$ be an irreducible polynomial in $K$. Then, the following assertions hold:
- $L := K[X]/\langle f \rangle$ is an extension of the field $K$;
- $f$ has a zero $\alpha$ in $L$, and we have $K(\alpha) \cong L$;
- $[L : K] = \textrm{deg} f$.
The first claim is easy to prove: $\langle f \rangle$ is a maximal ideal, since $K[X]$ is a PID, so $K[X]/\langle f \rangle$ is a field and $\phi: K \to K[X]/\langle f \rangle$ defined by $\phi(a) = a + \langle f \rangle$ is a homomorphism of fields, therefore also a monomorphism.
For the first part of 2, we note that $\alpha = X + \langle f \rangle \in L$ is a zero of $f$ in $L[X]$. Now, for $K(\alpha) \cong L$, my notes say that $\Phi: K(\alpha) \to L$, $\Phi(p(\alpha)) = p + \langle f \rangle$ is an isomorphism. ($K(\alpha) = K[\alpha]$ is obvious, since $\alpha$ is algebraic over $K$)
However, I don't know how to prove that $\Phi$ is well-defined: if $a = p_{1}(\alpha) = p_{2}(\alpha) \in K[\alpha]$, $p_{1}, p_{2} \in K[X]$, how do I prove that $p_{1} + \langle f \rangle = p_{2} + \langle f \rangle$, i.e. that $f | p_{1} - p_{2}$?
If I understand well what causes a problem to you, you can say that $(p_1-p_2)(\alpha)=0$, so, as $f$ is irreducible (and is therefore the minimal polynomial of $\alpha$), $f$ divides $p_1-p_2$, in other words $p_1\equiv p_2\mod f$.