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I have a task where I need to find a polynomial $R \in \mathbb{Q}[X]$ that has roots $\alpha + \beta$ where $\alpha$ are arbitrary roots of $X^5 - 2$ and $\beta$ of $X^4 - 3\enspace.$

The general technique to use is obvious to me, the polynomial is given by $$ R = \prod_{i, j} (x - (\alpha_i + \beta_j)) $$ with the corresponding roots $$ \begin{align*} \alpha_1 &= \sqrt[5]{2}\\ \alpha_2 &= -\sqrt[5]{-2}\\ \alpha_3 &= (-1)^{\frac{2}{5}}\sqrt[5]{2}\\ \alpha_4 &= -(-1)^{\frac{3}{5}}\sqrt[5]{2}\\ \alpha_5 &= (-1)^{\frac{4}{5}}\sqrt[5]{2}\\ \beta_1 &= \sqrt[4]{3}\\ \beta_2 &= -\sqrt[4]{3}\\ \beta_3 &= i\sqrt[4]{3}\\ \beta_4 &= -i\sqrt[4]{3}\enspace. \end{align*} $$ (compare to questions like Find polynomial whose root is sum of roots of other polynomials)


Now I know that one can argue that $R$ is indeed in $\mathbb{Q}[X]$, but we didn't yet step into Galois-theory so I probably can't use that, nor do I know the details of how to prove this.

The problem with manually computing $R$ is that the resulting equation is huge, really huge. After some hours of Mathematica magic I could resolve the above to $$ \begin{align*} S &= X^{20} - 15X^{16} -8X^{15} + 90X^{12} -1560X^{11} + 24X^{10} - 270X^8 - 11160X^7\\ &\qquad- 4080X^6 - 32X^5 + 405X^4 - 7560X^3 + 3960X^2 - 480X - 227\enspace. \end{align*} $$

The problem is that, if I go from this direction, I would need to check all $5 \cdot 4 = 20$ roots. But that is again a huge task.


At this point I'm stuck, both directions aren't eligible for a manual approach. I probably need some kind of fancy argument with which I can follow that $R = S$ or that $S$ has those roots (despite the fact that $S$ fall out of the sky).

Alternatively, arguments that $R$ is in $\mathbb{Q}[X]$ due to construction. But in this case I would need some guidance because, as said, I'm not sufficiently acquainted with Galois-theory.

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    $\begingroup$ It's quite easy to write down a $20$-by-$20$ matrix whose characteristic polynomial is your polynomial. Do you like computing characteristic polynomials of large matrices? $\endgroup$ – Lord Shark the Unknown Dec 15 '17 at 3:31
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    $\begingroup$ After some hours of Mathematica magic ... Or in WA resultant[ x^5-2, (z-x)^4 - 3, x ], or minimalpolynomial[ surd(2,5)+surd(3,4), z ]. $\endgroup$ – dxiv Dec 15 '17 at 3:44
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    $\begingroup$ I thijk that you need to know just about symmetric polynomials (which would be MUCH less than Galois Theory). $\endgroup$ – Wlod AA Dec 15 '17 at 3:55
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I'm not sure if this this is the argument you want. To be precise, what I want to explain is that for monic irreducible polynomials $f(x)$ and $g(x)$, and their roots $\alpha_i$, $\beta_j$ are the conjugates of the roots of them, respectively, the polynomial $$ p(x) = \prod_i \prod_j \left( x-\alpha_i -\beta_j \right) $$ is in $\mathbb{Z}[x]$ without referring the Galois theory.

Viewed as a polynomial over $\mathbb{Z}[\alpha_1, \alpha_2, \ldots , \alpha_n]$, the coefficients of $p(x)$ are symmetric polynomials of $\beta_j$s, say $\sigma_k$ be the degree $k$ elementary symmetric polynomial of $\beta_j$s. Therefore coefficients of $p(x)$ are of the form $B \left( \sigma_1 , \sigma_2 , \ldots , \sigma_m , \alpha_1 , \alpha_2 , \ldots , \alpha_n \right) $. Hence $B$ is a polynomial of $\alpha_i$s with integral coefficients. However, $B$ is also symmetric of $\alpha_i$. Therefore, is an integer. This proves that $p(x)$ is in $\mathbb{Z}[x]$.

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