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In another question, someone discussed about the fact that, if $p_1,\ldots,p_n$ is a list of integer, then the polynomial $$f=\prod_{e_1,\ldots,e_n\in\{\pm1\}}(x+e_1\sqrt{p_1}+\cdots+e_n\sqrt{p_n})$$ has integer coefficient. Can someone explain better this to me?

edit: the link of the previous question is Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$, and you have to multiple factor for every possible choice of $e_1,\ldots,e_n\in\{\pm1\}$

Lios
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  • linking to the actual question would be very helpful – gt6989b Apr 10 '18 at 19:27
  • The claim is clearly false as it is now...With $;n=1;$ , $;x+\sqrt2;$ is a counter example, or even $;(x+\sqrt2)(x+\sqrt3)=x^2+(\sqrt2+\sqrt3)x+\sqrt6;$ ... – DonAntonio Apr 10 '18 at 19:32
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    @DonAntonio I believe the OP means to multiply over all possible choices of signs. If so, it is surely correct for $n=1$, say. – lulu Apr 10 '18 at 19:32
  • @lulu That could be, yet the OP must then clarify this. – DonAntonio Apr 10 '18 at 19:33
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    @DonAntonio Oh, agreed. Worth noting: under my interpretation, the expression is obviously fixed under all elements of $Gal(\mathbb Q(\sqrt {p_1},\cdots, \sqrt{p_n})/\mathbb Q)$. – lulu Apr 10 '18 at 19:35
  • Consider the polynomial $p(x,y_1,y_2,...,y_n)=\prod(x+e_1y_1+...+e_ny_n)$. Then $p(x,y_1,y_2,...,y_n)=p(x,e_1y_1,...,e_ny_n)$ for all $e_i\in{\pm1}$. therefore, all $y_i$ appear to even powers in $p$. In fact, $\frac{1}{2}(p(x,y_1,...,y_n)+p(x,y_1,...,-y_n))=p(x,y_1,...,y_n)$ doesn't have terms in which $y_n$ appear to an odd exponent. Evaluating at $y_i=\sqrt{p_i}$ you get your result. –  Apr 10 '18 at 19:48
  • Sorry, I edited the post to make it more clear. – Lios Apr 10 '18 at 19:59

3 Answers3

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Hint:   by induction, using nothing more than the identity $\,(a+b)(a-b)=a^2-b^2\,$:

$$ \prod_{e_1,\ldots,e_n\in\pm1}(x+e_1\sqrt{p_1}+\cdots+e_n\sqrt{p_n}) = \prod_{e_1,\ldots,e_{\color{red}{n-1}}\in\pm1} \big((x+e_1\sqrt{p_1}+\cdots+e_{n-1}\sqrt{p_{n-1}}+\sqrt{p_n})(x+e_1\sqrt{p_1}+\cdots+e_{n-1}\sqrt{p_{n-1}}-\sqrt{p_n})\big) = \prod_{e_1,\ldots,e_{n-1}\in\pm1} \big((x+e_1\sqrt{p_1}+\cdots+e_{n-1}\sqrt{p_{n-1}})^2 - p_n\big) $$

Repeat the same in order to prove that the base of the squared term has integer coefficients.

dxiv
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The polynomial $$p(x,y_1,y_2,...,y_n)=\prod_{e_i\in\{\pm1\}}(x+e_1y_1+...+e_ny_n)$$

satisfies $p(x,y_1,...,y_n)=p(x,e_1y_1,...,e_ny_n)$ for all $e_i\in\{\pm1\}$.

Therefore, $$p(x,y_1,...,y_n)=\frac{1}{2}\left(p(x,y_1,...,y_i,...,y_n)+p(x,y_1,...,-y_i,...,y_n)\right)$$

doesn't have terms in which $y_i$ appears to odd powers, since in the right-hand side all such terms are cancelled.

The coefficients of $P(x)=p(x,y_1,...,y_n)$ are polynomials in $y_1,...,y_n$ with integer coefficients. This is clear from the definition of $p$, since the coefficients are just sums and products of $\pm y_i$.

Therefore, $f(x)=p(x,\sqrt{p_1},...,\sqrt{p_n})$ has integer coefficients.

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I assume that the question is: let $f\in \mathbb{Q}[x]$ such that $f$ has the given factorization in $\mathbb{C}[x]$. Show that $f\in \mathbb{Z}[x]$. Or maybe you are considering the product over all possible signs, which will be in $\mathbb{Q}[x]$ (explained in the comments). In any case, you can argue as follows:

The roots of $f$, which are those linear combinations of (plus minus) square roots of primes, are integral elements over $\mathbb{Z}$, because a sum of integral elements is integral. You may assume that your polynomial $f$ is irreducible (if not, split it in irreducible factors and argue the same for each of them). But then, you are looking at the minimal polynomial of those integral elements. By definition, this minimal polynomial must have coefficients in $\mathbb{Z}$.

Pedro
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  • It is right, but I have to find that this polynomial has integer coefficient to show that is the minimal polynomial of the sum (plus minus) of square roots of primes. So I'm looking for a proof that doesn't involve the fact that is the minimal polynomial. – Lios Apr 10 '18 at 19:57
  • A polynomial is the minimal polynomial for $x$ if it is monic, irreducbile and has $x$ as a root. So the irreducible factors of $f$ (which are monic) are trivially the minimal polynomials of their roots. Their roots are integral over $\mathbb{Z}$, so the coefficients are in $\mathbb{Z}$. So the irreducible factors of $f$ are in $\mathbb{Z}[x]$, and hence so is $f$ itself. – Pedro Apr 10 '18 at 20:04
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    This assumes a priori that $f$ has coefficients in $\mathbb Q$ though. – Carmeister Apr 10 '18 at 23:16
  • No it doesn't. I said: if the product is only over a subset of possible signs, then you need to assume it (otherwise its false, e.g. take a single factor); and if the product is over all possible signs, then the polynomial has rational coefficients. The reason is that you are including all Galois conjugates as roots of your polynomial (there cannot be a Galois conjugate which isn't of that form). This implies that the irreducible factors are minimal polynomials of some of them, and therefore have rational coefficients. But thanks for the remark, I guess I didn't make it clear enough. I'll edit – Pedro Apr 11 '18 at 05:23