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:

  1. $L := K[X]/\langle f \rangle$ is an extension of the field $K$;
  2. $f$ has a zero $\alpha$ in $L$, and we have $K(\alpha) \cong L$;
  3. $[L : K] = \textrm{deg} f$.

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 $f | p_{1} - p_{2}$?

  • $\begingroup$ I think it would be easier to construct an isomorphism in the other direction. $\endgroup$ – asdq Feb 15 '18 at 21:41

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 $p_1\equiv p_2\mod f$.

  • $\begingroup$ Yeah, I just noticed that myself - the key part that was bugging me was proving that $f$ is the minimal polynomial of $\alpha$ over $K$. Or at least $f_{1}$, which is $f$ scaled so that it's monic. However, this was actually trivial to prove. Thanks for confirming my "suspicion"! $\endgroup$ – Matija Sreckovic Feb 15 '18 at 21:44

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