Why totally positive elements of a number field have a root

Question

I heard that in a number field $K$

(let say $K = \mathbb{Q}[X]/P$), the totally positive element (that are sent into $\mathbb{R}^+$ for all morphisms of fields $K->\mathbb{C}$) all have a squareroot in $K \otimes_\mathbb{Q} \mathbb{R}$ (so, $\mathbb{R}[X]/P$).

In other words, : they are on the form f^2 for $f \in K \otimes_\mathbb{Q} \mathbb{R}$.

Would you have a reference that make the demonstration of this claim?

Thanks for your help!

Edit: my question wasn't precise enough and contained some mistakes I apology for it. I added some precisions.

Answer 1

It's quite clearly false if you mean that it has a square root in $K$: for instance $2\in \mathbb{Q}$ is obviously totally positive, but has no square root in $\mathbb{Q}$.

Or course any element of $K$ has a square root in a bigger number field $K\subset L$, but that's not very interesting and has nothing to do with positivity.

So in the end I'm not sure what statement you are referring to, it's possible you are confused.