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Dog’s group theory

Posted by teageegeepea under

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I liked John Derbyshire’s “Unknown Quantity: A Real and Imaginary History of Algebra”, but I still didn’t really grok why fifth order (and higher) polynomials don’t have general solutions. The answer has to do with group theory, which I don’t think is taught in high-school. For a while I had it in mind to eventually learn some group theory, but didn’t act on it. Last night I decided to try googling for an introduction, and one of the first results was, of all things, a tripod page. I know, who uses them. Seems pretty good though.

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March 15, 2011 at 7:32 am

I am not an expert or anything, but I think the basic idea goes like this:

a field is basically any set where you can add, multiply, and divide all numbers (except 0). The rational numbers (fractions) are a field, the real numbers, the complex numbers, etc.

These examples imply that you can have smaller fields “sitting inside” bigger ones. One way to construct fields from smaller fields is to take an already existing field, throw in a new element, and then throw in the minimum number of new elements to make this new set a field.

e.g. take the rationals, but add the square root of 2. Then, you will need to add all numbers of the form a*sqrt(2) where a is rational and * means multiplication, since a field is closed under multiplication. Also, all numbers of the form a + b*sqrt(2), since a field is closed under addition as well. It turns out that this is enough; further multiplication, say, of a + b*sqrt(2) and c + d*sqrt(2), yields a*c + 2*b*d + sqrt(2)*(a*d + b*c); and you can divide a + b*sqrt(2)/[c + d*sqrt(2)] by multiplying top and bottom by c – d*sqrt(2), and reduce it to a case of division.

So, you have made a new field from an old one by throwing in a square root. You can also make new fields by throwing in roots of polynomial equations.

So, we have two ways of making new fields out of the rationals: we can add radicals, or we can add roots of polynomials.

In each case, we can ask how the old field sits inside the new one. For example, in the case where we throw in sqrt(2), we also have to throw in -sqrt(2). The group theory here is looking at what group we get that interchanges the elements of the new field, while leaving elements of the old field alone. In the case of sqrt(2), this is just the group that interchanges sqrt(2) with -sqrt(2); the group of permutations of 2 elements.

In the case of a field generated by adding roots of polynomials, the group will in general be the group of permutations of n elements, where n is the degree of the polynomial.

The proof of impossibility basically says that the groups we get by starting with the rationals and adding in radicals can never add up to being the group of permutations of n elements for n greater than 4. It is impossible to have those smaller groups sit inside the 5-element permutation group in such a way that we can build the 5-element permutation group by successively throwing in new radicals into a field and looking at the symmetry group we get.

Does that make sense?

March 15, 2011 at 7:33 am

John Baez is always a good source for simple explanations of complicated things.

March 15, 2011 at 3:25 pm

I attempted Unknown Quantity last summer and found it even harder than Prime Obsession, for which I could at least skim over some of the more complicated parts while still getting the basic idea. Maybe it’s a good idea to supplement with outside sources while reading.

March 15, 2011 at 6:50 pm

Dog’s page doesn’t seem to get into fields. It does discuss multiplication if zero is removed from the set of numbers, because there is no way to “undo” a multiply-by-zero.

I don’t have Derbyshire book with me, I read it a while back. The library doesn’t have a copy, but I’m too lazy to pick it up now. Still have to finish “War in Human Civilization”!

March 15, 2011 at 7:39 pm

I only glanced at that page, but it seems to be more about group theory in general, rather than the specific group theory you need to understand why you can’t solve quintics via radicals. But yeah, the idea behind a field is that it is a group both additively AND multiplicatively (except 0, which has no multiplicative inverse). Division is just multiplication by inverses.

Man, I know what it’s like to not finish books…

March 16, 2011 at 8:02 am

_Abstract Algebra and Solution by Radicals_, reprinted by Dover for ten bucks or so, is a pretty good standalone description of how this works. It seems to be about as elementary as possible while still being serious: targeted roughly at a smart motivated college freshman.