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"For instance, the Galois group immediately tells you whether a polynomial can be solved at all ..." I wish she wrote one more sentence before changing the sub

"For instance, the Galois group immediately tells you whether a polynomial can be solved at all ..." I wish she wrote one more sentence before changing the subject. If it's immediate, Can anyone explain the decision procedure?

gizmo686 5y ago
A polynomial is solvable if and only if its corresponding Galois group is "solvable". Solvable groups are motivating primarily by the problem of determining if polynomials are solvable. In general, it takes some work to know if a group is solvable, and I don't think there is a way to determine that immidiatly. The most charitable reading of the statement is that there are relativly few groups of small order. This means that for polynomials of relativly small degree you have probably seen their Galois group before so can go off of memory.
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 ago
  The 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.
IngoBlechschmid 5y ago
The big picture is the following: There is a certain kind of groups called "solvable groups". There is a theorem stating that a polynomial equation can be solved by radicals if and only if its Galois group is a solvable group. And whether a given finite group (such as a Galois group) is solvable is an entirely finite combinatorial question which can be mechanically checked. Even better: A witness for the solvability of the Galois group can be transformed -- again nontrivially but mechanically -- into explicit formulas for the solutions of the polynomial equation. Wikipedia has some examples on solvable groups and their connection to Galois theory, though I hope that someone has written some blog post or similar explaining this beautiful connection with less prerequisites. https://en.wikipedia.org/wiki/Solvable_group https://en.wikipedia.org/wiki/Solvable_group

gizmo686 5y ago

A polynomial is solvable if and only if its corresponding Galois group is "solvable". Solvable groups are motivating primarily by the problem of determining if polynomials are solvable. In general, it takes some work to know if a group is solvable, and I don't think there is a way to determine that immidiatly. The most charitable reading of the statement is that there are relativly few groups of small order. This means that for polynomials of relativly small degree you have probably seen their Galois group before so can go off of memory.

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 ago
The 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.

IngoBlechschmid 5y ago

The big picture is the following: There is a certain kind of groups called "solvable groups". There is a theorem stating that a polynomial equation can be solved by radicals if and only if its Galois group is a solvable group. And whether a given finite group (such as a Galois group) is solvable is an entirely finite combinatorial question which can be mechanically checked. Even better: A witness for the solvability of the Galois group can be transformed -- again nontrivially but mechanically -- into explicit formulas for the solutions of the polynomial equation. Wikipedia has some examples on solvable groups and their connection to Galois theory, though I hope that someone has written some blog post or similar explaining this beautiful connection with less prerequisites. https://en.wikipedia.org/wiki/Solvable_group https://en.wikipedia.org/wiki/Solvable_group