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This question already has an answer here:

I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if $\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ (linear combination of the basis elements for $\mathbb{Q}(\sqrt{3}, \sqrt{2})$). I'm just not sure what.

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marked as duplicate by Ross Millikan, José Carlos Santos, Daniel W. Farlow, JonMark Perry, Dando18 Jul 25 '17 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/questions/1713529/… $\endgroup$ – Bill Cook Jul 25 '17 at 2:53
  • $\begingroup$ You can go with the top answer's route via solving equations or with Jyrki's answer where he uses the Galois correspondence. Without using the Galois correspondence, I think this is just a messy slog through a system of equations. $\endgroup$ – Bill Cook Jul 25 '17 at 2:54
  • $\begingroup$ Do you know Kummer theory? $\endgroup$ – Lord Shark the Unknown Jul 25 '17 at 5:32
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See Bill Dubuque's proof here and Iurie Boreico's proof here. See also [1], [2], [3], [4], [5].

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You can show that the polynomial $x^2-5$ is irreducible over $\Bbb{Q}[\sqrt{2},\sqrt{3}]$ by considering $x^2-5\bmod 23$ and noting that $23$ splits completely in $\Bbb{Q}[\sqrt{2},\sqrt{3}]$.

Edit: Let $K=\Bbb{Q}[\sqrt{2},\sqrt{3}]$ and $O_K$ be the ring of integers of $K$. If we knew $x^2-5$ was irreducible over $K$ we'd know $K[\sqrt{5}:K]=2$. What we'd like to find is a prime $p$ for which $x^2-5$ is irreducible over $\Bbb{F}_p$ and $p$ splits completely in $O_K$. Then for a $\frak{p}$ in $O_K$ lying over $p$ we'll have $x^2-5$ is irreducible over $O_{K}/\frak{p}$, hence irreducible over $K$. By trial and error using the Legendre symbol $\left(\frac{5}{23}\right)=-1, \left(\frac{2}{23}\right)=1,\text{ and} \left(\frac{3}{23}\right)=1$, so $23$ is a prime which satisfies all the requirements.

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  • $\begingroup$ I would upvote this answer if it explained in a little more detail. $\endgroup$ – Alfred Yerger Jul 25 '17 at 3:42
  • $\begingroup$ @AlfredYerger Just trying to keep it "fast." Probably too fast though. $\endgroup$ – sharding4 Jul 25 '17 at 4:02

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