So I am trying to show that $f^2,$ a non constant polynomial, cannot divide $x^n - 1.$ Suppose it did. Then $x^n - 1 = f^2g.$ The derivative of $x^n - 1$ is $nx^{n - 1} = 2ff'g + f^2g'.$ Clearly, $f^2$ and $2ff'g + f^2g$ share a factor of $f.$ However, $x^n - 1$ and $nx^{n - 1}$ are coprime. Contradiction. Does this work?
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Why are $x^n - 1$ and $n x^{n-1}$ are coprime? Note that this is as polynomials, the ring is Euclidean. – Will Jagy May 01 '17 at 01:55
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What could go wrong? – Mariano Suárez-Álvarez May 01 '17 at 01:55
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1@OP Idea is right, but the write-up would be improved if you pointed out where specifically you used the premise that $f$ is non constant (which you impliclity do, but don't have it spelled out). – dxiv May 01 '17 at 02:00
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@Mar the Op does not really know why the two polynomials $x^n-1$ and $n x^{n-1}$ are coprime. I blame you. – Will Jagy May 01 '17 at 02:25
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Polynomials are coprime in the ring if we can solve $fh_1 + gh_2 = 1, $ nonzero constant polynomial. And, you see, $$ \frac{x}{n} n x^{n-1} - (x^n - 1) = 1 $$
Apparently a confusing point, the field of coefficients can be taken to be the reals, for example. Then the coefficient of $x$ in the polynomial $\frac{x}{n}$ is the field element (constant) $\frac{1}{n}.$
Will Jagy
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But don't we need $\frac{x}{n}$ to exist in the polynomial ring? How do we know that $\frac{x}{n}$ necessarily exists? – green frog May 01 '17 at 02:03
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@ntntnt thought so. The ring $F[x]$ is euclidean when $F$ is a field. In this case, we may take, for example, $F = \mathbb R$ – Will Jagy May 01 '17 at 02:05
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It does work. A couple places I would add detail. You want $f^2$ to be the square of a non constant polynomial. As written, $f(x)$ could be $\sqrt {x-1}$, so $f^2=x-1,$ a non-constant polynomial and $(x-1)|(x^n-1)$. Maybe there is some context that says $f(x)$ must be a polynomial, in which case this is not needed. Then $nx^{n-1}$ and $x^n-1$ are not necessarily coprime, but the common factor must divide $x[nx^{n-1}]-n(x^n-1)=n$. As $f$ is non-constant it cannot always divide a constant.
Ross Millikan
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