MathNugget
Junior Member
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- Feb 1, 2024
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Reference for radical extensions: https://en.wikipedia.org/wiki/Radical_extension#:~:text=Solvability by radicals,-Radical extensions occur&text=In fact a solution in,a radical extension of K.
Short summary: a field extension K/k is simple radical if K=k[[imath]\alpha[/imath]], and [imath]\alpha^n=a[/imath], for some [imath]a \in k[/imath] and an [imath]n \in \mathbb{Q}[/imath] (probably definition works over a bigger field, but let's make the problem easier).
A (general) radical extension is a set of simple radical extensions [imath]F_1\subset F_2\subset F_3...[/imath], with each [imath]F_{i-1}\subset F_i[/imath] simple radical...
Well, we first break the polynomial up a little: [imath]f(x)=3x^6-5x^5+x^2=x^2(3x^4-5x^3+1)[/imath]. As [imath]x^2[/imath] provides nothing to the extension, I am left with a 4th degree polynomial. According to the internet, Abel–Ruffini theorem says "if the degree of the polynomial is less or equal to 4, it's definitely solvable by radicals". I guess this completes the proof? the field extension is now generated by some radicals...
Short summary: a field extension K/k is simple radical if K=k[[imath]\alpha[/imath]], and [imath]\alpha^n=a[/imath], for some [imath]a \in k[/imath] and an [imath]n \in \mathbb{Q}[/imath] (probably definition works over a bigger field, but let's make the problem easier).
A (general) radical extension is a set of simple radical extensions [imath]F_1\subset F_2\subset F_3...[/imath], with each [imath]F_{i-1}\subset F_i[/imath] simple radical...
Well, we first break the polynomial up a little: [imath]f(x)=3x^6-5x^5+x^2=x^2(3x^4-5x^3+1)[/imath]. As [imath]x^2[/imath] provides nothing to the extension, I am left with a 4th degree polynomial. According to the internet, Abel–Ruffini theorem says "if the degree of the polynomial is less or equal to 4, it's definitely solvable by radicals". I guess this completes the proof? the field extension is now generated by some radicals...
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