# Closure of quadratic extensions under taking powers

I'll give a proof of the following.

THEOREM: Let $x = p+\sqrt{q}$, with $p,q \in \mathbb{Q}$, and $m$ an integer. Then $x^m = a+b\sqrt{q}$ with $a,b \in \mathbb{Q}$.

PROOF

Using the Binomial Theorem

$${x^m} = C_0^m{p^m} + C_1^m{p^{m - 1}}{q^{\frac{1}{2}}} + C_2^m{p^{m - 2}}{q^{\frac{2}{2}}} + \cdots + C_{m - 2}^m{p^2}{q^{\frac{{m - 2}}{2}}} + C_{m - 1}^mp{q^{\frac{{m - 1}}{2}}} + C_m^m{q^{\frac{m}{2}}}$$

Let $m= 2·j$ then

$${x^{2j}} = C_0^{2j}{p^{2j}} + C_1^{2j}{p^{2j - 1}}{q^{\frac{1}{2}}} + C_2^{2j}{p^{2j - 2}}{q^{\frac{2}{2}}} + \cdots + C_{2j - 2}^{2j}{p^2}{q^{\frac{{2j - 2}}{2}}} + C_{2j - 1}^{2j}p{q^{\frac{{2j - 1}}{2}}} + C_{2j}^{2j}{q^{\frac{{2j}}{2}}}$$

Grouping produces

$${x^{2j}} = \sum\limits_{k = 0}^j {C_{2k}^{2j}{p^{2j - 2k}}{q^k}} + \sum\limits_{k = 1}^j {C_{2k - 1}^{2j}{p^{2j - 2k + 1}}{q^{k - 1}}\sqrt q }$$

But since every binomial coefficient is integer, and every power of $p$ and $q$ is rational then one has

$${x^{2j}} = a+b\sqrt{q} \text{ ; and } a,b \in \mathbb{Q}$$ where $b$ and $a$ are the sums.

If $m = 2j+1$ then

$$x^{2j+1} =(a+b\sqrt{q}) (p+\sqrt{q}) = c+d\sqrt{q}$$

which is also in our set.

(I don't know if the ring-theory tag is appropiate.)

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To answer your question: I don't see any error. Moreover - the last step surprised/pleased me after I was just starting to think about the case of odd m. Very nice that detail.... –  Gottfried Helms Feb 6 '12 at 17:17
I would throw in a few parentheses in the final expression for $x^{2j}$, but fine. Also, there seems to be no need to separate even and odd case: just group the terms with even exponent for $q$ and those with odd exponent for $q$. Finally: you can use the argument you used for odd $m$ to prove this with a simple inductive argument: it's true for $m=1$ and $m=2$, and if it is true for $k$, then it is true for $k+1$ (since it holds for $m=2$). –  Arturo Magidin Feb 6 '12 at 17:18
Right. I decided to use an even exponent to make things simpler. I guess induction would make things look a little less messy. (I mean, my last step really suggests the use of induction) –  Pedro Tamaroff Feb 6 '12 at 17:21
Would this set be called $$\mathbb{Q}[\sqrt{q}]$$ or something of the sort? –  Pedro Tamaroff Feb 6 '12 at 17:26
@Peter: Yes, or equivalently $\mathbb{Q}(\sqrt{q})$ since $\sqrt{q}$ is algebraic over $\mathbb{Q}$. –  Clive Newstead Feb 6 '12 at 17:27

It seems correct, but it's much longer-winded than it needs to be. You could do this easily by induction and without any need for the binomial theorem:

Let $a,b,c,d \in \mathbb{Q}$.

If we can write $$(a+b\sqrt{q})(c+d\sqrt{q}) = r+s\sqrt{q} \qquad (*)$$ for some $r,s \in \mathbb{Q}$, then we're done by induction.

Why? Because this will tell you that any finite product of elements of the form $a+b\sqrt{q}$ will take the same form, and a special case of this is a positive integer power of an element of said form, e.g. $(p+\sqrt{q})^m$.

So expanding the left-hand side of $(*)$ tells you that choosing $$r = ac + bdq\ ,\ s = ad+bc$$ works since $q$ is rational.

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Yes. That is what my las comment to Arturo suggested. Thanks. –  Pedro Tamaroff Feb 6 '12 at 17:24

More conceptually and more generally, suppose $\alpha$ is the root of a monic polynomial over a ring $R$

$$\alpha^n\ =\ r_{n-1}\ \alpha^{n-1} +\:\cdots\:+r_1\ \alpha + r_0$$

Multiplying the above by $\alpha$, and using the above as a rewrite rule to replace $\alpha^n$ by the lower powers on the RHS, we deduce that $\alpha^{n+1}$ may also be expressed as on the RHS. By induction so too can every power of $\alpha$, hence $R[\alpha] = R + \alpha\ R +\:\cdots\:+ \alpha^{n-1}\ R$, i.e. every polynomial in $\alpha$ with coefficients in $R$ is equal to one of degree $< n$.

Equivalently, if monic $g(\alpha)= 0$ for $g(x)\in R[x]$ of degree $n$, then, by the Division Algorithm, any polynomial $f(\alpha)$ is equal to an $h(\alpha)$ of degree $< n$, where $h(x)$ is the remainder of $f(x)$ mod $g(x)$

$$f(x)\ =\ q(x)\ g(x) + h(x)\ \ \Rightarrow\ \ f(\alpha)\ =\ h(\alpha)\ \ by\ \ g(\alpha) = 0$$

Recall that the high-school long Division (with Remainder) Algorithm works for any monic polynomial over any ring, so the the above normal-form degree-reduction algorithm works for any element $\alpha$ that is a root of monic polynomial. Such $\alpha$ generalize pure radicals $\sqrt[n]{r},\ r\in R$ and are known as algebraic integers over $R$, a fundamental concept in number theory and algebra.

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Thanks. I guess I have to get into algebra a little more. –  Pedro Tamaroff Feb 6 '12 at 18:58