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2

It's enough to show that $\mathbb{C}((X))$ has an $n$-th root of $1+X$. One way to do this is using the binomial theorem to expand $(1+X)^{1/n}$.


4

Note that $x^6+1=(x^3+1)^2=(x-1)^2(x^2+x+1)^2$. The polynomial $x^2+x+1$ is irreducible over $\mathbb{F_2}$ and hence $\mathbb{F_2}/(x^2+x+1)\cong\mathbb{F_4}$ is an extension field of $\mathbb{F_2}$. In that field $x^2+x+1$ has a root (the element $x+(x^2+x+1)$ is a root), and since $\mathbb{F_4}/\mathbb{F_2}$ is a normal extension all roots of $x^2+x+1$ ...


1

Hint: Over ${\Bbb F}_2$: $x^6+1 = (x^3+1)^2$ by freshman's dream.


1

You are correct. $x^7-2\in \mathbb{Q}[x]$ is irreducible of degree $7$. On the other hand, the "obvious" polynomial $x^7-1\in \mathbb{Q}[x]$ for the $7^{th}$ roots of unity already has a root in $\mathbb{Q}$, namely $1$. So, if we factor this out, we get $$ x^7-1=(x-1)\underbrace{(1+x+x^2+x^3+x^4+x^5+x^6)}_{\text{irreducible of degree}\:6}.$$ So, $[\mathbb{...


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