When does $\sqrt{a_1} + \cdots + \sqrt{a_k} \in K$ imply $\sqrt{a_1}, \ldots, \sqrt{a_k} \in K$?

Question

Consider $R = \mathbb{Z}[X_1, \ldots, X_k]$, the polynomial ring in $k$ variables over $\mathbb{Z}$, and let $S = \mathbb{Z}[\sqrt{X_1}, \ldots, \sqrt{X_k}]$. Then $S/R$ is an integral extension of commutative ring; let $f \in R[T]$ be the minimal polynomial of $\alpha = \sqrt{X_1} + \cdots + \sqrt{X_k} \in S$. One sees (for example by passing to the fraction fields and invoking Galois theory) that $\deg(f) = 2^k$.

Some examples of $f$ for small values of $k$:

• $k = 1$: $f = T^2 - X_1$
• $k = 2$: $f = T^4 - 2(X_1 + X_2)T^2 - (X_1 - X_2)^2$
• $k = 3$: $ f =T^8 + (-4(X_1 + X_2 + X_3))T^6 + (6(X_1^2 + X_2^2 + X_3^2) + 4(X_1X_2 + X_1X_3 + X_2X_3))T^4 + (-4(X_1^3 + X_2^3 + X_3^3) + 4(X_1^2X_2 + X_1^2X_3 + X_2^2X_1 + X_2^2X_3 + X_3^2X_1 + X_3^2X_2) - 40X_1X_2X_3)T^2 + (X_1^2 + X_2^2 + X_3^2 - 2(X_1X_2 + X_1X_3 + X_2X_3)^2$

Now consider some arbitrary field $K$ of characteristic different from $2$ and some fixed homomorphism $\varphi : R \to K$. I want to know when the existence of a root of $f(\varphi(X_1), \ldots, \varphi(X_k), T)$ in $K$ implies that $\varphi(X_1), \ldots, \varphi(X_K)$ are squares in $K$, i.e. that $\varphi$ extends to a homomorphism $S \to K$. My understanding is that this will somehow 'generically' be the case by the fact that $K(\sqrt{X_1}, \ldots, \sqrt{X_k}) = K(\sqrt{X_1} + \cdots + \sqrt{X_k})$ for any field $K$ of characteristic different from $2$, but one needs to exclude some values of $t$, possibly depending on the $\varphi(X_i)$. For example, if $k = 2$, $\varphi : R \to K$ any morphism and $t \in K$ such that $f(\varphi(X_1), \varphi(X_2), t) = 0$, then one can verify the identities $$ \varphi(X_1) = \left(\frac{t^2 - \varphi(X_2) + \varphi(X_1)}{2t}\right)^2 \enspace \text{and} \enspace \varphi(X_2) = \left(\frac{t^2 - \varphi(X_1) + \varphi(X_1)}{2t}\right)^2 $$ under the assumption that $t \neq 0$. Hence, any morphism $\varphi : R \to K$ such that there exists a $t \in K^\times$ with $f(\varphi(X_1), \varphi(X_2), t) = 0$ extends to a morphism $S \to K$. The condition $t \neq 0$ cannot be dropped: consider for example $\varphi : R \to \mathbb{R}$ defined by $\varphi(X_1) = \varphi(X_2) = -1$, then $0$ is a root of $f(-1, -1, T) = T^4 + 4T^2$, but $-1$ is not a square in $\mathbb{R}$.

How should I approach the problem of finding the values of $t$ one has to exclude for larger values of $k$ - say, for starters, for $k = 3$? Any pointers towards books or articles on related subjects are very welcome.

Answer 1

No hope, even for $k=3$ and for an arbitrary firld $K$ conaining a non-square (is it non-quadratically-closed?).

Let $K$ be such field and $\alpha\in K$ be a non-square; set $L=K[\sqrt\alpha]$. Take any $t\in K$ and define $\psi\colon S\to L$ by $\psi(\sqrt{X_1})=t$, $\psi(\sqrt{X_2})=\sqrt\alpha$, and $\psi(\sqrt{X_3})=-\sqrt\alpha$. Then the image of $\varphi=\psi\big|_R$ is contained in $K$, and $\psi(\sqrt{X_1}+\sqrt{X_2}+\sqrt{X_3})=t\in K$, so $t$ satisfies the required equation; but $\sqrt\alpha\notin K$

It seems that, in order to have a real hope, the weakest property you need is the $\sqrt{X_i}$ be linearly independent over $\varphi(R)$, or at least smewhat close to that (as a model example over $\mathbb Q$ shows). And this is hard to reach just by imposing some condition on $t$.