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Not sure what this statement is supposed to mean. Every polynomial can be solved tautologically by adding a zero for it into the picture. For instance, every co
by tacomonstrous 5y ago
Not sure what this statement is supposed to mean. Every polynomial can be solved tautologically by adding a zero for it into the picture. For instance, every complex polynomial admits a complex zero (this is the Fundamental theorem of algebra), so you can find the zeros of any rational polynomial somewhere. You can of course look at the smallest 'field' in which these zeros live: basically, look at everything you get by writing down all possible arithmetic expressions involving those zeros. For the polynomial x^2-2, you get numbers of the form a+\sqrt(2)b where a and b are rational numbers.
The Galois group tells you how complicated this new field is, and how involved its symmetries are. In the example above there is only one nontrivial symmetry which switches a+sqrt(2)b with a-sqrt(2)b. In general there are many more, and when you consider the symmetries of the zeros of all possible rational polynomials, you get the absolute Galois group of the rational numbers. It can be said without exaggeration that number theory is the study of this Galois group, which is still a rather mysterious object in its full glory.
- xyzzyz 5y agoThe intended (and, in fact, widely understood within mathematical community) meaning of that statement is whether the roots of polynomials can be expressed using radicals, ie. as algebraic expressions involving four standard arithmetic operations and taking roots. I’d rather quibble with the part where they say that this is “immediately” seen from Galois group, as showing solvability of a group is not trivial matter at all.