Let $L/K$ be an arbitrary algebraic field extension. How is a normal closure of $L$ (the smallest normal extension of $K$ containing $L$) constructed?

If $L/K$ is finite, then writing $L=K(\alpha_1,...,\alpha_n)$ the splitting field of the product of the minimum polynomials $\prod_i m_{K,\alpha_i}(X)$ is such a field, unique up to isomorphism.

Is there a method for its construction in the general case using a Zorn's lemma type argument?

Many thanks.

  • 2
    $\begingroup$ If you know there is an algebraic closure, you can take the intersection of all normal extensions of $K$ containing $L$ in this closure. Of course you can also construct it directly in a similar way to the algebraic closure (instead of adding roots of all polynomials, you only add the remaining roots of polynomials who already have a root). $\endgroup$ – Captain Lama Apr 13 '16 at 17:29
  • 2
    $\begingroup$ Agreeing with @CaptainLama, I’d say that for infinite extensions, you’ll usually need to appeal to Zorn, and say that if an overfield of $L$ is not yet normal, you find an element in it with a $K$-conjugate not in it. Adjoin that element. There is your induction step. For explicitly-given extensions, maybe the set of all real $n$-th roots of all primes, over $\Bbb Q$, there may well be techniques peculiar to your situation (here, adjoin as well all roots of unity). $\endgroup$ – Lubin Apr 13 '16 at 17:39
  • $\begingroup$ @Captain Lama Thanks but I am wondering if there is a way to construct it using a Zorn's lemma type argument (i.e. define a partially ordered set S, show that an arbitrary chain in S has a maximal element etc). $\endgroup$ – eddie Apr 13 '16 at 17:46
  • $\begingroup$ You just have to be careful because to apply Zorn's lemma directly you need a big ambiant extension on which to define your partial order. $\endgroup$ – Captain Lama Apr 13 '16 at 17:48
  • $\begingroup$ What if I choose an algebraic closure of $L$? regards $\endgroup$ – eddie Apr 13 '16 at 17:52

Your Answer

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

Browse other questions tagged or ask your own question.