1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I think it would be easier to construct an isomorphism in the other direction. $\endgroup$ – asdq Feb 15 '18 at 21:41
3
$\begingroup$

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 $p_1\equiv p_2\mod f$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.