Prove that $\Bbb Z[\sqrt{2}, \sqrt{3}]$ does not equal $\Bbb Z[\sqrt{2} + \sqrt{3}]$.

Question

Prove that $\Bbb Z[\sqrt{2}, \sqrt{3}]$ does not equal $\Bbb Z[\sqrt{2} + \sqrt{3}]$.

This is homework. I want to prove that these are different sets. The first set is the smallest ring containing the integers and the two radicals separately. The second is the smallest ring containing the integers and the sum of the two radicals. Unfortunately I cannot find any similar proofs to learn from. I've been trying to show that the second set does not contain the square roof of $6$, for instance, or the square root of $3$. We haven't looked at what the second set might contain. We've only looked at adjoining a single radical and not the sum of radicals on class. If someone could solve a similar example or give me a crucial insight, that would be great.

I suppose that root 2 is not in the second set, since that would imply that root 3 was and thus that the sets are equal. But it's not clear to me how to prove this. — OLP, Jan 18 '15 at 02:23
Honestly, no. We simply have not been taught how to do so. Of course I know that all sums and products are involved. But I don't know how to shape the chaos. With just root 2 adjoined, I could do so. Playing around with sums and products, it seems that one cannot isolate root 2 or root 3 or root 6. But I can't formalize this, only conjecture. — OLP, Jan 18 '15 at 02:44
Well, you seem to know what the elements of $\mathbb Q(\sqrt2+\sqrt3)$ are... Maybe a similar idea works? — Mariano Suárez-Álvarez, Jan 18 '15 at 02:49
Let's start with: what is the definition of the second ring? — Mariano Suárez-Álvarez, Jan 18 '15 at 02:51
Surely, both sets contain $\sqrt2+\sqrt3$ and all its powers. I suggest that you write out the first few powers of this number, simplifying appropriately, of course. — Lubin, Jan 18 '15 at 02:54

score 4 · Accepted Answer · answered Jan 18 '15 at 03:14

4

Hint $\ $ Note $\,\ \sqrt{3}\,\not\in \Bbb Z[\alpha]\ $ for $\ \alpha\, =\, \sqrt 3 +\! \sqrt{2}\ $ of degree $\,4\,$ over $\Bbb Q,\,$ else $$\!\!\!\! \begin{eqnarray} \alpha\,(2\sqrt 3-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt 3+\!\sqrt 2)(\sqrt 3-\!\sqrt 2)\, =\, \color{#0a0}{1}\\ \Rightarrow\ \ \alpha\sqrt 3\, =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{1}}2\,\in\,\color{}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$

answered Jan 18 '15 at 03:14

Bill Dubuque

272,048

The same argument works for $,\sqrt{a}+\sqrt{b},\ $ see this answer. – Bill Dubuque Jan 18 '15 at 03:18
We haven't talked about degrees, but this proof is terse and gives me the kind of contradiction I was hoping for. Thanks. – OLP Jan 18 '15 at 03:35
@s.z. There is further discussion in the comments of the linked generalization which may prove helpful in that regard. – Bill Dubuque Jan 18 '15 at 03:41
For your description of the ring you need more than knowing that a is of degree four over the rationals. – Mariano Suárez-Álvarez Jan 18 '15 at 03:48
3

This is quite violently unmotivated. – Mariano Suárez-Álvarez Jan 18 '15 at 03:48
@Mariano It's a hint / sketch. As I remarked, if the OP needs further details they are in the linked comments - which address your first remark. – Bill Dubuque Jan 18 '15 at 03:55
1

But why would anyone compute $a(2\sqrt3-a)$? – Mariano Suárez-Álvarez Jan 18 '15 at 04:21
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@MarianoSuárez-Alvarez: Conjugating $\sqrt{3} + \sqrt{2}$ seems like a natural thing to do, and then trying to express the conjugate $\sqrt{3} - \sqrt{2}$ as a $\mathbb{Z}$-linear combination of $\sqrt{3}$ and $\alpha$ makes sense. Solving $\sqrt{3} - \sqrt{2} = m\alpha + n\sqrt{3}$ yields $m=-1$, $n=2$. – Jan 18 '15 at 04:29
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@Bungo, and the answer would go from an opaque verification to an instructive example on how to deal with these things if it explain that. That is my point. – Mariano Suárez-Álvarez Jan 18 '15 at 04:33
@MarianoSuárez-Alvarez: I agree with your point. This answer (even as a hint) is a bit cryptic as written. – Jan 18 '15 at 04:40
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@Bungo Alas, I don't think there is a nice way to adequately explain what is going on here without employing more advanced concepts from algebraic number theory. – Bill Dubuque Jan 18 '15 at 04:51

user2345215 · Answer 2 · 2015-01-18T03:04:09.740

3

Not the most elegant solution, but note that $$(\sqrt2+\sqrt3)^{2n+1}=a_n\sqrt2+b_n\sqrt3,$$ where $a_0=b_0=1$ and $$(\sqrt2+\sqrt3)^2(a_n\sqrt2+b_n\sqrt3)=(5+\sqrt6)(a_n\sqrt2+b_n\sqrt3)=(5a_n+6b_n)\sqrt2+(5b_n+4a_n)\sqrt3,$$ so $$2\mid a_{n+1}-b_{n+1}\iff 2\mid(5a_n+6b_n)-(5b_n+4a_n)=a_n-b_n+2b_n.$$

Therefore the difference $a_n-b_n$ is even for all $n$. Even powers $(\sqrt2+\sqrt3)^{2n}$ contain just $\sqrt6$. But the elements of $\mathbb Z[\sqrt2+\sqrt3]$ have the form $$\sum_{k=0}^d z_k(\sqrt2+\sqrt3)^k.$$ This means the difference of multiples of $\sqrt2$ and $\sqrt3$ always has to be even, which contradicts $\sqrt2$ or $\sqrt3$ being elements of this ring.

I obviously assume we already know $\sqrt2, \sqrt3$ and $\sqrt6$ are linearly independent over $\mathbb Q$.

edited Jan 18 '15 at 03:04

answered Jan 18 '15 at 02:53

user2345215

16,422

Right you are. I find nothing inelegant in this. – Lubin Jan 18 '15 at 02:55
@Lubin I'm sure there's some slick argument why the difference is even and I missed it. – user2345215 Jan 18 '15 at 03:01
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Write down the first, second, and third powers of $\sqrt2+\sqrt3$ and observe that the fourth power is a $\Bbb Z$-linear conmbination of $1$ and these three numbers. – Lubin Jan 18 '15 at 03:04
Thanks for your help. We haven't talked out things being "over Q" yet. We've just learned the definition of IDs and fields. The approach to these added elements was just a sketch. – OLP Jan 18 '15 at 03:37

Prove that $\Bbb Z[\sqrt{2}, \sqrt{3}]$ does not equal $\Bbb Z[\sqrt{2} + \sqrt{3}]$.

2 Answers2