Link to Keith Conrad's paper describing the cubic resolvent and discriminants and how these relate to the Galois group: kconrad.math.uconn.edu/blurbs... Hope you enjoy! :)
Thanks very much! I think one reason is because the topics are niche. Only so many people who want to learn number theory/galois theory/etc. Hoping to get more in the future though
Hello, I believe the cubic resolvent should be R_3(x) = x^3 -4bx + a^2. I’m getting this definition from page 308 of algebra by Papantonopoulpou. Please let my know your thoughts.
I wasn't able to find a copy of that book easily accessible. I'm guessing the author just has a different definition of cubic resolvent. The one I'm using should be fine for computing Galois groups though.
In general the formula for the discriminant is given by the product formula I wrote, but if one wants to find the discriminant of a quartic polynomial which has the specific form X^4+aX+b then the discriminant is given by -27a^4+256b^3. It's difficult to show this, but you can do so by expanding the discriminant (again using the product formula from 7:20) and then using the fact that the resulting expression is a symmetric polynomial in the roots of f, see for example this article: math.stackexchange.com/questions/3699434/using-symmetric-polynomials-to-find-the-discriminant-of-x4-px-q-over-ma?rq=1
Excuse me, but I think the Galois group of x⁴ - 4x + 2 cannot be the S(4). S(4) has 24 operations, this means, ALL permutations of the 4 roots of this polynomial are automorphisms of this field extension, wich is impossible.
Good question. If you try to use p=2 the polynomial becomes x^3, which is reducible. If you reduce mod 3 you get x^3+x+2, which is reducible because 2 is a root. I ended up with 5 because that's the next smallest choice that gives you something irreducible.
Thanks so much!
I didn't expect you to have only this much subs with such great explanations
Thanks very much! I think one reason is because the topics are niche. Only so many people who want to learn number theory/galois theory/etc. Hoping to get more in the future though
Nice video on Galois group. By the way which software tool you used for the whiteboard.
I appreciate it! I use microsoft whiteboard (which can be slow but is generally reliable).
Hello, I believe the cubic resolvent should be R_3(x) = x^3 -4bx + a^2. I’m getting this definition from page 308 of algebra by Papantonopoulpou. Please let my know your thoughts.
I wasn't able to find a copy of that book easily accessible. I'm guessing the author just has a different definition of cubic resolvent. The one I'm using should be fine for computing Galois groups though.
Hi, i have a quick question: are u using Microsoft Whiteboard?
Are u using a stylus? Because the handwriting is great...
Thanks! This was done with microsoft whiteboard. I am also using a wacom tablet since otherwise writing stuff is impossible haha
7.20: what exactly is now "this specific formula"?
In general the formula for the discriminant is given by the product formula I wrote, but if one wants to find the discriminant of a quartic polynomial which has the specific form X^4+aX+b then the discriminant is given by -27a^4+256b^3.
It's difficult to show this, but you can do so by expanding the discriminant (again using the product formula from 7:20) and then using the fact that the resulting expression is a symmetric polynomial in the roots of f, see for example this article: math.stackexchange.com/questions/3699434/using-symmetric-polynomials-to-find-the-discriminant-of-x4-px-q-over-ma?rq=1
Excuse me, but I think the Galois group of x⁴ - 4x + 2 cannot be the S(4).
S(4) has 24 operations, this means, ALL permutations of the 4 roots of this polynomial are automorphisms of this field extension, wich is impossible.
Why did you chose 5 when using Z/5Z?
Good question. If you try to use p=2 the polynomial becomes x^3, which is reducible. If you reduce mod 3 you get x^3+x+2, which is reducible because 2 is a root. I ended up with 5 because that's the next smallest choice that gives you something irreducible.
You > Galois group
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