I'm trying to do/understand the following exercise:

"Let $E$ be a finite extension of a field $F$. If $[E:F] = 2$, show that $E$ is a splitting field of $F$."*

Background: Just beginning studying fields, I know nothing of Galois theory. Here is my reasoning so far:

$[E:F] = 2$ means that E, when viewed as a vector space over F, has dimension 2. This means that

$\beta \in E \implies \beta = b_0 + b_1 \alpha$, where $\alpha \in E \setminus F$

In other words, $E$ must not only be a finite extension but a simple extension $F(\alpha)$, where the degree of $\alpha$ over F is 2. Did I interpret the definitions correctly here? Does $[E:F]$ imply that $E$ is such a simple extension? [Cleared up definition of a field splitting a field.]

Now, assuming the above is correct, $\alpha \in E$ implies that $\alpha^2 = c+d\alpha$, for some $c$, $d \in F$. Consider this polynomial of degree 2 in $F[x]$:


We see that:

$p(\alpha)=\alpha^2-d\alpha-c = 0$


$p(x) = (x-\alpha)q(x)$

Where $q(x)$ must be of degree one, i.e. a linear factor. So $E$ is a splitting field of $p(x)$. Is my reasoning correct?

*pg. 330 from Judson's Abstract Algebra: Theory and Applications, freely available here http://abstract.ups.edu/download.html.

  • $\begingroup$ It means that $E$ splits some (irreducible) polynomial. $\endgroup$ – André Nicolas Mar 18 '14 at 0:39
  • $\begingroup$ Oh thank god. I will see if I can solve it from there. This may be a too meta of a question, but is it better form to leave my question as is or edit it now that I know that definition properly? $\endgroup$ – mb7744 Mar 18 '14 at 0:40
  • $\begingroup$ You can write down the polynomial, since $\alpha^2$ is a linear combination of $1$ and $\alpha$. $\endgroup$ – André Nicolas Mar 18 '14 at 0:42
  • $\begingroup$ Hard to know, perhaps yes. You will be given a solution, but you need not look at it until you have figured it out on your own. $\endgroup$ – André Nicolas Mar 18 '14 at 0:45
  • $\begingroup$ Sorry I am not sure what you meant by that last comment. I elected to alter my question because that definition isn't really the crux of the problem. Thank you for your hint though! $\endgroup$ – mb7744 Mar 18 '14 at 1:12

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