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I am seeking resolution of a disagreement over the nature of Galois theory. In the followup comments on a separate thread, I made the claim that the beauty of Galois theory is that it does not require us to distinguish between the various choices for square roots and and cube roots. Another correspondent disagreed; replying to my example, he wrote:

"You would be unable to even define "the Galois structure of x^10−1=0" if radicals did not define functions."

By which I am understanding him to say that in order to do Galois theory, we must have a convention whereby one choice the radical is distinguished as special.

I would claim that Galois theory is precisely about all the things you can say about algebraic structures without having to make any arbitrary choices as to which particular values of a radical we are talking about. To be sure, there are sometimes choices made, but they are never arbitrary. There is no need to have a single-valued function for the square root or the cube root. I wonder if people will agree with this characterization.

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I'm not sure what "Galois structure" means. Does this mean the Galois group? –  Qiaochu Yuan Sep 4 '11 at 5:12
    
I guess so. To me it means everything we do when we talk about permutations of roots and the group structrue. –  Marty Green Sep 4 '11 at 5:37
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I think you have to make a distinction between an abstract field and its embeddings. The notion of positive root only makes sense when you are considering an embedding in an ordered field like $(\mathbb{R},\leq)$. This is irrelevant to "basic" Galois theory. –  YBL Sep 4 '11 at 14:26

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up vote 4 down vote accepted

There is a sense in which it is not just unnecessary, but impossible, to distinguish between positive and negative square roots. E.g., in the field ${\bf Q}[x]/(x^2-2)$, is $x+(x^2-2)$ the positive square root of 2, or is it the negative?

On the other hand, sometimes it's useful to make the distinction. An easy way to show that $i$ isn't in $K={\bf Q}(\sqrt{1+\sqrt2})$ is to see that $K$ is contained in the reals, which you couldn't do if you didn't distinguish between $\sqrt2$ and $-\sqrt2$.

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