There are two problems in Fraleigh's text on abstract algebra.

Which are true?

1.$\mathbb{R}$ is a splitting field over $\mathbb{R}$
2.$\mathbb{R}$ is a splitting field over $\mathbb{Q}$

I could not understand the questions. I know the definition of a splitting field of a polynomial over some field but here, no polynomial is mentioned.

Can someone help to understand the problem and the process of solving these types of problem?

  • 2
    $\begingroup$ What does being a splitting field over another field mean? It is either a splitting field for a set of polynomials over some field or else a normal extension.... $\endgroup$ – DonAntonio Jun 27 '13 at 10:30


  1. Can you think of polynomial with coefficient in $\mathbb{R}$ such that all of its zeros are in $\mathbb{R}$ as well?
  2. The splitting field $L$ of a set of polynomials in $K[x]$ is gotten by adjoining all the zeros of those polynomials. Consequently $L/K$ is an algebraic extension.
  • $\begingroup$ But what is the case here.No plolynomial is mentioned here.I am still totally confused $\endgroup$ – ghotan Jun 27 '13 at 10:39
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    $\begingroup$ So I interpret the phrase "a splitting field over $K$" to mean "the splitting field over $K$ of some set of polynomials in $K[x]$." The task for you is to describe such a set, or prove that no such set exists. $\endgroup$ – Jyrki Lahtonen Jun 27 '13 at 10:44
  • $\begingroup$ @JyrkiLahtonen and ghotan: if you followed my comments below the other answer you know then that Fraleigh considers fields to be candidates for splitting fields over another field $\,k\,$ only those that are contained in $\,\overline k\,$...this automatically rules out $\,\Bbb R/\Bbb Q\,$....weird indeed! $\endgroup$ – DonAntonio Jun 27 '13 at 10:54
  • $\begingroup$ @DonAntonio: I don't find that weird. See my second hint. A splitting field is always algebraic over the base field, for me at least. $\endgroup$ – Jyrki Lahtonen Jun 27 '13 at 10:59
  • 1
    $\begingroup$ The whole concept from Fraleigh is what is odd for me, @JyrkiLahtonen . Of course, the request to be contained in the alg. closure of the base field (i.e., the extension is algebraic) makes sense. $\endgroup$ – DonAntonio Jun 27 '13 at 11:01

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