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Peter Mueller
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Consider $a$ as a transcendental over $\mathbb C$. Then theeach $x_i$'s generate a Galois generates an extension of $\mathbb C(a)$ of degree $n$ withwhose Galois closure has Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Consider $a$ as a transcendental over $\mathbb C$. Then the $x_i$'s generate a Galois extension of $\mathbb C(a)$ of degree $n$ with Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Consider $a$ as a transcendental over $\mathbb C$. Then each $x_i$ generates an extension of $\mathbb C(a)$ of degree $n$ whose Galois closure has Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Corrected the answer (added missing Galois group)
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Peter Mueller
  • 22.5k
  • 1
  • 75
  • 107

Consider $a$ as a transcendental over $\mathbb C$. Then the $x_i$'s generate a Galois extension of $\mathbb C(a)$ of degree $n$, thus with Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Consider $a$ as a transcendental over $\mathbb C$. Then the $x_i$'s generate a Galois extension of $\mathbb C(a)$ of degree $n$, thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Consider $a$ as a transcendental over $\mathbb C$. Then the $x_i$'s generate a Galois extension of $\mathbb C(a)$ of degree $n$ with Galois group the symmetric group $S_n$. Thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.

Source Link
Peter Mueller
  • 22.5k
  • 1
  • 75
  • 107

Consider $a$ as a transcendental over $\mathbb C$. Then the $x_i$'s generate a Galois extension of $\mathbb C(a)$ of degree $n$, thus once $n\ge5$, you cannot solve for the $x_i$'s in terms of $a$ by radicals.