Finding the degree of minimal polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:02:59Z http://mathoverflow.net/feeds/question/61985 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials Finding the degree of minimal polynomials Victor 2011-04-17T02:11:43Z 2011-04-17T18:52:00Z <p>Let a number $x = \sqrt[a_1]{p_1} + \sqrt[a_2]{p_2} + \ .. \ + \sqrt[a_n]{p_n}$ be a number such that all $a_n$ are integers and all $p_n$ are rational. I've been noticing that for every number x, the degree of its minimal polynomial is seemingly always equal to $\prod_{1}^n \ a_n$.</p> <p>Is that valid for all values of $a_n$? If so, is there a proof?</p> http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/61990#61990 Answer by The Cheese Stands Alone for Finding the degree of minimal polynomials The Cheese Stands Alone 2011-04-17T02:37:11Z 2011-04-17T18:48:24Z <p>No. Some conditions are needed on the $a_i$ and $p_i$. For instance, take n=2, $a_1 = a_2 = 2$, $p_1 = p_2 = 2$. Then $x = 2 \sqrt{2}$, which has minimal polynomial $x^2 - 8$. As an even simpler example, n=1, $a_1 = 2$, $p_1 = 4$, then $x$ is rational.</p> <p>For a less trivial example, take $a_1= 4$, $a_2 = 6$, $p_1=p_2=2$. Check that this has a polynomial of degree 12. In fact, this isn't really true at all. </p> <p>One can, however, prove that the degree of the minimal polynomial is at most $\prod a_n$, which is an easy exercise in field theory. Any graduate algebra textbook covering Galois theory will be more than sufficient to prove this; just remember the degree of the minimal polynomial is the same as the dimension of the extension field viewed as a vector space over the base field.</p> <p>EDIT:</p> <p>After much miscommunication on my part, we've reached the following results:</p> <p>Suppose $a_1,\ldots,a_n$ are pairwise relatively prime positive integers, $p_1, \ldots, p_n$ integers such that $\sqrt[a_i]{p_i}$ is of degree $a_i$ for each i. Then $\sqrt[a_1]{p_1} + \cdots + \sqrt[a_n]{p_n}$ is of degree $\displaystyle \prod_{i=1}^n a_i$.</p> <p>The condition that each $\sqrt[a_i]{p_i}$ is met (by Eisenstein Criterion) should there be a prime $q_i$ such that $q_i | p_i$ and $q_i^2 \not{|} p_i$ for each i.</p> http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/62024#62024 Answer by Georges Elencwajg for Finding the degree of minimal polynomials Georges Elencwajg 2011-04-17T11:45:19Z 2011-04-17T18:52:00Z <p>Besicovitch has proved the following related interesting result:</p> <blockquote> <p>Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ has dimension $n^k$ over $\mathbb Q$ .<br> More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )$$ </p> </blockquote> <p>(The case $n=2$ is a classical chestnut in Galois theory.)<br> This does not answer the OP's question but at least assures us that, for example, $$\sqrt[3]{900}+\sqrt[3]{36}+ \sqrt[3]{15}+\sqrt[3]{150} \notin \mathbb Q$$<br> which is not so simple to check directly. </p> <p>I have the pessimistic feeling that there is no very satisfactory general answer to the question "when does the sum $\sqrt[n_1]{a_1}+ \sqrt[n_2]{a_2}+...+\sqrt[n_k]{a_k}$ have degree $n_1 n_2 ...n_k$", but I'd love to be shown wrong.</p> <p><strong>Bibliography:</strong> Besicovich's original article is: Abram S. Besicovitch, "On the linear independence of fractional powers of integers", Journal of the London Mathematical Society 15 (1940), 3-6.</p> <p>Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273. 13 (1974), 268-273.</p> http://mathoverflow.net/questions/61985/finding-the-degree-of-minimal-polynomials/62049#62049 Answer by Igor Rivin for Finding the degree of minimal polynomials Igor Rivin 2011-04-17T18:13:40Z 2011-04-17T18:13:40Z <p>The canonical references for this are:</p> <p>MR0818878 (87b:68058) Zippel, Richard(1-MIT-C) Simplification of expressions involving radicals. J. Symbolic Comput. 1 (1985), no. 2, 189–210. </p> <p>MR1148819 (92k:12008) Landau, Susan(1-MA-C) Simplification of nested radicals. SIAM J. Comput. 21 (1992), no. 1, 85–110. </p> <p>and more recently</p> <p>MR1776235 (2001g:12004) Blömer, J.(D-PDRB) Denesting by bounded degree radicals. (English summary) Fifth European Symposium on Algorithms (Graz, 1997). Algorithmica 28 (2000), no. 1, 2–15. </p>