# $\mathbb{Q}[\sqrt a_1,\ldots,\sqrt a_k]$ vs. $\mathbb{Q}(\sqrt a_1,\ldots,\sqrt a_k)$

Call an algebraic number polyquadratic if it can be expressed as the sum or difference of a finite number of square roots of rational numbers. (This definition follows Conway-Radin-Sadun rather than Rotman.)

Is the inverse of a nonzero polyquadratic number polyquadratic? I thought not, but so far I have not been able to find an example. I have proved (on the literal back of an envelope, so rather informally) that sums of 1 through 4 square roots are indeed invertible as polyquadratic numbers. Does this pattern continue or was my initial intuition right?

Your question essentially amounts to asking whether $\mathbb{Q}[\sqrt{a_1},\ldots,\sqrt{a_k}]$ is a field. This is a special case of the following lemma:
$\textbf{Lemma:}$ Let $A$ be an integral domain which is finite dimensional over a subfield $K$. Then $A$ is a field.
$\textit{Proof:}$ To prove the lemma, suppose $x\in A$ is nonzero and look at the linear map $\ell_x:A\rightarrow A$ defined by $\ell_x(y)=xy$. This is injective because $x$ is not a zero divisor. An injective linear transformation from a finite dimensional vector space to itself is surjective, so we can find $y$ such that $xy=\ell_x(y)=1$. Hence every nonzero element is a unit.
Setting $A=\mathbb{Q}[\sqrt{a_1},\ldots,\sqrt{a_k}]$ and $K=\mathbb{Q}$ shows that the inverse of a nonzero polyquadratic number is polyquadratic.
Alternatively, one can use the fact that any polyquadratic number $\alpha$ is algebraic. In particular there is a polynomial equation of the form $\alpha^n+a_{n-1}\alpha^{n-1}+\ldots+a_0=0$ with each $a_i$ rational. Furthermore $a_0\neq 0$ so $1=\alpha a_0^{-1}(\alpha^{n-1}+a_{n-1}\alpha^{n-2}+\ldots+a_1)$, which makes $a_0^{-1}(\alpha^{n-1}+a_{n-1}\alpha^{n-2}+\ldots+a_1)$ the inverse to the polyquadratic number $\alpha$. Since polyquadratic numbers form a ring, this inverse is polyquadratic as well.