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I'm trying to solve an exercise as it follows:

$\alpha, \beta \in \mathbb{C}$ such that $a^{37}=2=\beta^{17}$. Note that both are prime.

a) Find the minimum polynomials of $\alpha$ and $\beta$ over $\mathbb{Q}$.

Well, I guess that $(x^{37}-2)$ and $(x^{17}-2)$ are irreducible over $\mathbb{Q}$, hence minimum polynomials. But does that follow from 37,17 being prime, or just from the fact that the polynomial has no zero in $\mathbb{Q}$? How do I write this down rigorously?

b) Compute $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}]$, and compute $[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$!

My guess here on the first one is that this is equal to $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\beta)}]*[\mathbb{Q}( \beta):\mathbb{Q}]$ and $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\alpha)}]*[\mathbb{Q}( \alpha):\mathbb{Q}]$, so the first product is $17x$, the second is $37y$, so the dimension of field extension is at least $17\times 37$, but this is also the maximum number (right?), so $17\times 37$ is the answer? and this would be the same answer as for the last part, $\mathbb{Q}(\alpha \beta)$? How do I write this down nicely?

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Eisenstein Criterion. Having no rational root does not imply irreducibility. –  André Nicolas Dec 8 '11 at 16:25
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Your title asks about finding minimum polynomials, but the question is about degrees of extension. You don't need to find minimum polynomials to determine the degree of the extension in many cases (as evidenced by your "guess", actually a correct proof) of the first part of question (b). For the second part, the degree of $\mathbb{Q}(\alpha\beta)$ over $\mathbb{Q}$, to get you started: since $\mathbb{Q}(\alpha\beta)\subseteq \mathbb{Q}(\alpha,\beta)$, the degree $[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$ must divide $17\times 37$; hence the degree is either $1$, $17$, $37$, or $17\times 37$. –  Arturo Magidin Dec 8 '11 at 16:50

2 Answers 2

up vote 2 down vote accepted

$\mathbb{Q}(\alpha,\beta)$ is the composite field of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$, which have relatively prime degrees 37 and 17. Hence, by a standard theorem on extensions, the degree of the composite is 37*17. You are right that the maximum possible degree of a composite is the product of the individual degrees. The minimum pssible degree is their least common multiple.

In regard to Andre's comment, the polynomial $x^{4}+4$ has no rational roots,but factors over the rationals as $(x^{2}+2x+2)(x^{2}-2x+2)$.

As for part 2 of your question (b), consider a polynomial in $\mathbb{Q}[x]$ and evaluate it at $\alpha\beta$. Can you relate this element to any of the extensions in your final paragraph?

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In what you are trying to do you can replaces the number 2 by any positive integer $a$ that is not a power. Then $x^{n}-a$ is irreducible over $\mathbb{Q}$ by the Vahlen-Cappelli criterion for irreducibility. You can replace 17 and 37 by any positive integers $m$ and $n$. If $\alpha$ is an $m$-th root of $a$ and $\beta$ is an $n$-th root of $a$, then it is not hard to show that $\mathbb{Q}(\alpha,\beta)$ has degree $lcm(m,n)$ over $\mathbb{Q}$. –  Chris Leary Dec 8 '11 at 17:26

Regarding the first part of your question, it's not true in general that if $f\in\mathbb{Q}[x]$ has no rational root then $f$ is irreducible. For example, consider $f(x)=(x^2-2)(x^2+1)$. However, if you are familiar with Eisenstein's criterion for irreducibility, you'll see that it can be used to show that $x^n-2$ is irreducible over $\mathbb{Q}$ for every $n>0$. So in fact this part doesn't depend on those two exponents being prime. However, the fact that they are prime will come into play in the second part of the problem.

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