Is there a splitting field for multivariate polynomials over $\mathbb{Q}$? Let $f(X,Y) = X^2 + Y^2 - c$.  Does there exist a field $F$ such that $f(X,Y) = \prod_{i=1}^n(a_1^i X + a_2^i Y + a_3^i)$, where $a^i$'s are in $F$?
 A: No. The zero locus of a product like
$$
\prod_{i=1}^n(a_1^iX+a_2^iY+a_3^i)
$$
is the union of the lines
$$
a_1^iX+a_2^iY+a_3^i=0.
$$
On the other hand the zero locus of $X^2+Y^2-a$ is a circle of radius $\sqrt a$. A circle is not a union of lines no matter how you turn the coordinates.

You can apply the theory of splitting fields and such, but then you can only have a single variable. The other variables must be included into the field!
So you can do the following. Let $K=\mathbb{Q}(X)$ be the field of rational functions of the variable $X$. You can view $f$ as a quadratic polynomial in the ring $K[Y]$. Then $f$ has a splitting field
$$K_f=K[\sqrt{c-X^2}],$$
i.e. you need to adjoin a square root of an element of $K$.
A: Yes, although probably not in the way you'd like it. If you put $F={\bf Q}(X,Y)$, then $f(X,Y)\in F$, and your condition is trivially satisfied.
For a more meaningful question, you could ask if there is a field $F$ of characteristic $0$ such that $X,Y$ are transcendent over $F$ and $f$ factors into linear terms over $F[X,Y]$. In which case, Jyrki's answer is probably the right way to see it.
A different way to show that it's not possible for $c\neq 0$ is by direct calculation, for example with two linear terms (other cases we can rule out by degree considerations):


*

*Suppose that it is otherwise and $X^2+Y^2-c=(a_1X+b_1Y+c_1)(a_2X+b_2Y+c_2)$. This means that $a_1a_2=b_1b_2=1,c_1c_2=-c$ etc.

*Without loss of generality $a_1=1$ (we can multiply the second linear term by $a_1$ and the first by $a_1^{-1}$). Then $a_2=a_1^{-1}=1$ (by examining the $X^2$  term).

*Furthermore if we look at $XY$ term, we get that $a_1b_2+b_1a_2=b_1+b_2=0$, and by $Y^2$ term $b_1\neq 0$.

*Now, we look at $X$ and $Y$ terms. They give us $a_1c_2+c_1a_2=c_1+c_2=0$ and $(b_1c_2+c_1b_2)=b_1(c_2-c_1)=0$. This implies that $2c_1=0$, so because characteristic is not $2$, $c_1=c_2=0$ and $c=0$. This is indeed possible, as we have $X^2+Y^2=(X+iY)(X-iY)$.
The same proof will work for any field of characteristic $\neq 2$. In characteristic $2$, we have $X^2+Y^2-c=(X+Y+\sqrt c)^2$.
