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I am currently studying the theory of field extensions.

One of the exercises in my book is causing me a trouble. The problem is:

Give an example of a field $E$ containing a proper subfield $K$ such that $E$ is embeddable in $K$ and $[E:K]$ is finite.

First, I could't find any example. To me, it seems like any field in typical textbook cannot be an example.

Second, it seems to me that the problem states that it is possible for two fields to be a field extension of the other but they are not the same(since $K$ is proper). Is it correct? I thought that field extension can be used to establish some partial order, if we restrict fields that are subfields of some given field (for example, algebraically closed field of $E$, in this case).

Any examples or hints would be highly appreciated.

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You need a field that is isomorphic to a proper subfield. Algebraic extensions of, say, ${\mathbb Q}$, won't work, so you need to resort to transcendental extensions. For instance: ${\mathbb Q}(X) : {\mathbb Q}(X^2)$.

Note that you can embed the "larger" field ${\mathbb Q}(X)$ in the "smaller" field ${\mathbb Q}(X^2)$ by sending $X$ to $X^2$. (So, the embedding doesn't have to be an inclusion.)

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  • $\begingroup$ And it looks like $[Q(X):Q(X^2)] = 2$; am I right? $\endgroup$ – Robert Lewis Apr 4 '14 at 5:38
  • $\begingroup$ Yes, indeed, the degree is 2; the minimum polynomial of $X$ over ${\mathbb Q}(X^2)$ is $Y^2 - X^2$. $\endgroup$ – Magdiragdag Apr 4 '14 at 5:41
  • $\begingroup$ Thanks for the clever answer! I gave a thought on that example but failed to prove that the extension is finite. $\endgroup$ – Scream Apr 4 '14 at 5:47
  • $\begingroup$ Magdiragdag, can I take $X$ to be $\pi$? $\endgroup$ – Biswarup Saha Sep 10 '18 at 9:37
  • $\begingroup$ @BiswarupSaha Yes, you can take $\pi$ instead of $X$, as $\pi$ is transcendental over ${\mathbb Q}$. This does mean that you'll be embedding ${\mathbb Q}(\pi)$ in ${\mathbb Q}(\pi^2)$ by sending $\pi$ to $\pi^2$; this may look more confusing than embedding ${\mathbb Q}(X)$ in ${\mathbb Q}(X^2)$ by sending $X$ to $X^2$, even though it is the same idea. $\endgroup$ – Magdiragdag Sep 10 '18 at 10:39

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