Do Groebner bases over rationals lose information about complex solutions? Given a system of polynomial equations, if I calculate the Groebner basis over the rationals and get {1} then:

*

*My understanding is that I can then conclude there are no rational solutions. Is this correct?


*Can I also conclude that there are no complex solutions as well? (Because there are no complex solutions to 1=0? ... or is that being too flippant?)


*If they do lose information on complex solutions, under what conditions does this occur? I've made some tests and haven't found a case where information is lost. So it appears, at the very least, there are many cases in which no information is lost. In particular, what if I restrict to systems of equations which are multilinear in the variables? (ie. it could have a fourth order term such as $wxyz$ but there is never a term with a variable raised to second or higher power)
Or is there no simple condition and I just need to recalculate the Groebner basis over the complex polynomial space?

I'm sorry if these questions sound very basic, but this is outside of the math I've learned and I used a CAS to help me solve a problem. So I want to be absolutely sure I am not misusing the tools and making some incorrect assumptions (and Mathematica only calculates Groebner basis over the rationals, yet makes claims about complex systems of polynomials, so I need to understand what I can trust).
Here is a simple test:
In[1]: GroebnerBasis[{x^2 + y, y - 1}, {x, y}, CoefficientDomain -> Rationals]
Out[1]: {-1 + y, 1 + x^2}

And even though there are no rational solutions, it did not reduce to {1}. So at least in this case, it did not lose information about complex solutions.
In the comments, Dietrich Burde suggested adding x^2 + y^2 + 1
In[2]: GroebnerBasis[{x^2 + y, y - 1, x^2 + y^2 + 1}, {x, y}, CoefficientDomain -> Rationals]
Out[2]: {1}

This collapsed to {1}, so this is more directly related to my question. In this case there are no complex solutions either, so no information was lost.
Here is a more involved example, with multilinear equations that only has rational solutions with $w=x=0$, but has complex solutions with all variables non-zero ($x=y=1$, $w=z=i$):
$$
\begin{aligned}
x y + w z &= 0 \\
-x z + y w &= 0 \\
\end{aligned}
$$
And here is what Mathematica calculates for the Groebner basis over the rationals
In[1]: GroebnerBasis[{x y + w z, -x z + y w}, {w, x, y, z}, 
 CoefficientDomain -> Rationals] 
Out[1]: {x y^2 + x z^2, x y + w z, w y - x z}

The non-trivial complex solution is still a solution, so it does not appear any information was lost here either. However the basis is no-longer multi-linear, which suggests multi-linear equations are not somehow special here.
 A: Consider the set of complex solutions to a system of polynomial equations $$S = \{(p_1,\ldots,p_n) \in \mathbb{C}^n \mid f_i(p_1,\ldots,p_n) = 0 \text{ for each } i=1,\ldots,r \}$$ where the $f_i \in \mathbb{Q}[x_1,\ldots,x_n]$ are polynomials in $x_1,\ldots,x_n$ with rational coefficients.
If a Groebner basis of the ideal $I = \langle f_1, \ldots, f_r \rangle \subset \mathbb{Q}[x_1,\ldots,x_n]$ contains $1$, then in particular $1$ is contained in the ideal $I$, which means $1 = \sum_{i=1}^r c_i f_i$ for some polynomials $c_i \in \mathbb{Q}[x_1,\ldots,x_n]$. Any solution $p = (p_1,\ldots,p_n)$ in $S$ is then also a solution to $1 = \sum_{i=1}^r c_i(p)f_i(p) = 0$ (i.e. $1=0$), so there are no solutions: $S = \varnothing$. In the same way there are no rational solutions either ($S \cap \mathbb{Q}^n = \varnothing$).
If (a Groebner basis of) $I$ does not contain a nonzero constant, then the system of equations has a complex solution: $S \neq \varnothing$ (this is Hilbert's weak Nullstellensatz). The question of existence of rational solutions (rational points) is generally more subtle (it is a whole field of study).
When you compute a Groebner basis of $I$ as an ideal in $\mathbb{Q}[x_1,\ldots,x_n]$ (e.g. using Buchberger's algorithm), then effectively you will only do plain arithmetic with the input polynomials $f_i$, like division with remainder, addition, subtraction, multiplication, and division by nonzero constants. In other words, all the arithmetic takes place in $\mathbb{Q}[x_1,\ldots,x_n]$, and there is never any need to go to a bigger ring, or bigger field of coefficients. If you consider $I$ as an ideal in $K[x_1,\ldots,x_n]$, for any field extension $K$ of $\mathbb{Q}$ (in particular for $K=\mathbb{C}$), and you run the same algorithm, then all the steps will be the same and so the end result will be the same (effectively, all computation happens in the subring $\mathbb{Q}[x_1,\ldots,x_n]$). That is, a Groebner basis of $I$ computed in $\mathbb{Q}[x_1,\ldots,x_n]$ will also be a Groebner basis of $I$ in $K[x_1,\ldots,x_n]$, in particular also for $K=\mathbb{C}$.
