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Motivated by the positive answer to the following question:

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

I'm curious about whether ${\Bbb Z}[\sqrt{2}+\sqrt{3}]=\Bbb{Z}[\sqrt{2},\sqrt{3}]$ is also true, where ${\Bbb Z}[\alpha]$ denotes the smallest subring of ${\Bbb C}$ which contains $\alpha\in{\Bbb C}$. It suffices to know whether $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ is true or not. With some manipulation one can get $$ 2\sqrt{2}\in {\Bbb Z}[\sqrt{2}+\sqrt{3}]. $$ It seems no hope to get $f(x)\in\Bbb{Z}[x]$ such that $$ f(\sqrt{2}+\sqrt{3})=\sqrt{2}, $$ which is equivalent to $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$. But I don't have a proof. How should I go on?

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Letting $\alpha = \sqrt{2} + \sqrt{3}$, we have $\sqrt{2} = \frac{1}{2}\alpha^3 - \frac{9}{2}\alpha \in \mathbf{Q}(\alpha)$.

The minimal polynomial of $\alpha$ over $\mathbf{Q}$ is $f(x) = x^4 - 10x^2 + 1$. Thus any element $z$ of $\mathbf{Q}(\alpha)$ has a unique representation of the form $z = a\alpha^3 + b\alpha^2 + c\alpha + d$ with $a, b, c, d \in \mathbf{Q}$. Moreover, because $f(x)$ has integer coefficients, we have $z \in \mathbf{Z}[\alpha]$ if and only if $a, b, c, d \in \mathbf{Z}$.

It follows that $\sqrt{2} \not\in \mathbf{Z}[\alpha]$.

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    $\begingroup$ Readers might wonder where $\,\sqrt{2}\, =\, \frac{1}{2}\alpha^3 - \frac{9}{2}\alpha\ $ comes from. Below is a derivation of this $$\alpha(\alpha^3\!-10\alpha)\, =\, -1\ \Rightarrow\ \color{#c00}{\alpha^{-1}} = 10\alpha-\alpha^3$$ $$2\sqrt{2}\, =\, (\sqrt{3}+\sqrt{2})-(\sqrt{3}-\sqrt{2})\, =\, \alpha-\color{#c00}{\alpha^{-1}} = \alpha^3 - 9\alpha$$ But there is a simpler one-line proof that avoids this - see my answer. $\endgroup$ Jan 17 '14 at 14:04
  • $\begingroup$ Why "because $f(x)$ has integer coefficients, we have $z\in{\Bbb Z}[\alpha]$ if and only if $a,b,c,d\in{\Bbb Z}$"? $\endgroup$
    – user9464
    Jan 17 '14 at 19:44
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    $\begingroup$ @Jack. When I wrote this answer, I should have written "because $f(x)$ has integer coefficients and $f(x)$ is monic." To prove the assertion, note that $\mathbf{Z}[\alpha]$ consists of all $h(\alpha)$ for polynomials $h(x) \in \mathbf{Z}[x]$. So it is enough to prove by induction that $\alpha^n$ is of the required form. This proves necessity. Sufficiency then follows from the uniqueness of $a, b, c, d$. $\endgroup$
    – user122171
    Jan 18 '14 at 6:59
  • $\begingroup$ Where does this come from? I would say: $$\sqrt 2 + \sqrt 3 = \alpha \implies \alpha^2 - 5 = 2\sqrt 6$$ and $$2\sqrt 6 \alpha = 4\sqrt 3 + 6 \sqrt 2 \implies (\alpha^2 - 5)\alpha - 4\alpha = \alpha^3-9\alpha = 2\sqrt 2.$$ $\endgroup$
    – Watson
    Feb 7 '17 at 12:55
  • $\begingroup$ This answer shows that a necessary and sufficient condition for the inclusion $\Bbb Z[a+b] \subseteq \Bbb Z[a,b]$ to be an equality, when $a,b$ are algebraic integers such that $\Bbb Q(a+b) = \Bbb Q(a,b)$ (which is a necessary condition for equality), is that when writing $a=g(a+b)$ in a unique way with $g \in \Bbb Q[X]$ of degree $< \deg_{\Bbb Q}(a+b)$, we have $g \in \Bbb Z[X]$. $\endgroup$
    – Watson
    Feb 7 '17 at 12:55
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Without being a 'pretty' solution, it's not too difficult to show that even powers of $\sqrt{2}+\sqrt{3}$ are of the form $a+b\sqrt{6}$ with $a,b\in {\Bbb Z}$, and odd powers are of the form $a\sqrt{2}+b\sqrt{3}$ with $a,b\in{\Bbb Z}$ and $a,b$ both odd.

No matter how many of the latter you add up, you'll never have the parity of the coefficients differing.

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  • $\begingroup$ To build off this, define $g : {\Bbb Z}[\sqrt2 + \sqrt3] \rightarrow \Bbb Z$ by $g(a\sqrt 2 + b\sqrt3 + c\sqrt 6) = a+b+2 c$. Then $g((\sqrt 2 + \sqrt 3)^n)$ is even for all $n$, hence for any $f\in \Bbb Z[x]$, it must be that $f(g(a))$ is even for all $a\in \Bbb Z$. In other words, use an appropriate 'norm' for this finitely generated $\Bbb Z$-module. $\endgroup$
    – gabe
    Jan 17 '14 at 0:27
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Hint $\ $ Note $\,\ \sqrt{b}\,\not\in \Bbb Z[\alpha]\ $ for $\ \alpha\, =\, \sqrt b +\! \sqrt{a}\ $ of degree $\,4\,$ over $\Bbb Q,\,$ else $$\!\!\!\! \begin{eqnarray} \alpha\,(2\sqrt b-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt b+\!\sqrt a)(\sqrt b-\!\sqrt a)\, =\, \color{#0a0}{b\!-\!a}\\ \Rightarrow\ \ \alpha\sqrt b\, =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{b\!-\!a}}2\,\in\,\color{#c00}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$

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    $\begingroup$ To show $\,\Bbb Z[\alpha] = {\Bbb Z}+\alpha{\Bbb Z}+\alpha^2{\Bbb Z}+\alpha^3{\Bbb Z} =: R\,$ it suffices to show $\,\alpha^4\in R,\,$ since this implies the same for all higher powers $\,\alpha^n.\,$ But $\,\alpha^4\in R\iff \alpha\,$ is a root of a monic quartic $\in\Bbb Z[x].\,$ This is true: square $\alpha^2 + \color{#0a0}{b\!-\!a}\, =\, 2\alpha\sqrt{b}\ $ from above. Alternatively, if you know about algebraic integers then $\,\alpha = \sqrt{a}+\sqrt{b}\,$ is the sum of two algebraic integers, so it too is an algebraic integer, therefore its monic minimal polynomial is $\in \Bbb Z[x].$ $\endgroup$ Jan 17 '14 at 22:20
  • $\begingroup$ The above was in reply to a question posed in a comment (since deleted). $\endgroup$ Jan 17 '14 at 22:22
  • $\begingroup$ Here is the deleted comment: I don't understand how to get ${\Bbb Z}[\alpha]\subset {\Bbb Z}+\alpha{\Bbb Z}+\alpha^2{\Bbb Z}+\alpha^3{\Bbb Z}$. $\endgroup$
    – user9464
    Jan 17 '14 at 22:25
  • $\begingroup$ @Jack For your other question, I presume $\alpha$ has degree $4$, so it cannot be the root of lower degree polynomial over $\Bbb Q.\,$ So two polynomials in $\alpha$ with coef's $\in \Bbb Q\,$ of degree $\le 3$ are equal iff they have equal coef's. In your case you can see this quickly since a nonzero poly of deg $\le 3$ cannot have the $4$ roots $\,\pm\sqrt{2}\pm\sqrt{3}.$ That those are all roots of said quartic follows because the coef's have form $\,p(a,b) = p(\sqrt{a}^2,\sqrt{b}^2)$ for $\,p\in\Bbb Z[x,y],\,$ so changing the signs of the sqrt's does not change the value of the quartic. $\endgroup$ Jan 18 '14 at 19:40

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