Galois extension of $K$ generated by the roots of polynomials $f(x_1,x_2, \cdots, x_n) \in K[x_1,x_2, \cdots, x_n]$. Let us consider a field $K$ of characteristic $0$. Then we know that any finite extension $L$ of $K$, which is a Galois extension as well, is produced the roots of a separable polynomial $f(x) \in K[x]$. This is know as Galois theory for one variable polynomial.
My question:
My question is about multivariable Galois theory i.e., Galois extension of $K$ generated by the roots of polynomials $f(x_1,x_2, \cdots, x_n) \in K[x_1,x_2, \cdots, x_n]$.
$(1)$ Can the roots of some $2$-variable polynomial $f(x_1,x_2) \in K[x_1,x_2]$ generate a Galois extension of $K$ ?
The following is more elaborated question.
$(2)$ Consider a $2$-tuple polynomial $f(X)=(f_1(x_1,x_2),f_2(x_1,x_2)) \in K[x_1,x_2]^2$, where $X=(x_1,x_2)$ is a $2$-tuple. Then the zeros of $f(X)$ are given by $f_1(x_1,x_2)=f_2(x_1,x_2)=0$ i.e., the common solutions. Let us define $$S=\{(\alpha_1,\alpha_2) \in L^2: ~f_1(\alpha_1,\alpha_2)=0=f_2(\alpha_1,\alpha) \},$$ $\text{where $L$ is some finite extension of $K$}$
From this post, I know that the set $S$ of solution can be algebraic and thus $L$ might be a Galois extension of $K$. However I am confused about the matter that ''the elements of $K$ are just scalars i.e., $1$-tuple while the elements in $S$ are like coordinates or $2$-tuple vector. So how does the solutions $f_1(x_1,x_2)=f_2(x_1,x_2)=0$ generate an algebraic extension of $K$ ?
Example:
As In my second question, suppose I consider $f(X)=(f_1(X),f_2(X)) \in K[X] \times K[x]$ by $f(x,y)=(x^2+y^2-2,x^2-2y^2-1)$, where $X=(x,y)$.
Then zeros of $f(X)$ are given by $x^2+y^2-2=0=x^2-2y^2-1$ and those are $(\pm \sqrt{\frac{5}{3}}, \pm \frac{1}{\sqrt{3}})$. In this case, how would we describe the extension ?
Is it like $K(\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}, \sqrt{\frac{1}{3}},-\sqrt{\frac{1}{3}}) \cong K(\sqrt{\frac{5}{3}})$
or like $K\left((\sqrt{\frac{5}{3}}, \sqrt{\frac{1}{3}}), (-\sqrt{\frac{5}{3}},\sqrt{\frac{1}{3}}), (\sqrt{\frac{5}{3}},-\sqrt{\frac{1}{3}}), (-\sqrt{\frac{5}{3}},-\sqrt{\frac{1}{3}})\right)$ ?
Clearly the first case is degree one extension. What about 2nd case ?
Any discussion please.
 A: Too long for a comment - but a reply to k-rational's question/comment... I hope of interest to Why, too.
Suppose one is 'given' $f_1,\cdots,f_m\in K[x_1,\cdots, x_n]$. Then the 'zero-set' $V\subset \mathbb A_K^n$ of these polys is also the zero-set of an ideal $\subset K[x_1,\cdots, x_n]$ - call the ideal $I$ - I'm  sweeping  (ideal generated by $f_1,\cdots,f_m$? radical ideal? variety? scheme?) details under the rug, enough, I hope, that what's left is sufficiently 'expansive' so as to leave no room for mistake...
In any case, $V$ 'corresponds' to a polynomial ring $A=K[x_1,\cdots, x_n]/I$, where $I$ is the 'defining ideal' of $V$; we say $A$ is the ring of functions on $V$. Fix a $K$-algebra $L$ (for instance, a field – or not - perhaps $L=A$). A point $P=(a_1,\cdots,a_n)\in V(L)$, (i.e., $a_i\in L$) defines a $K$-alg homomorphism $e_P:A \to L$, defined by $x_i\mapsto a_i$ - the letter 'e' in $e_P$ stands for 'evaluation: that is, as $x_i\mapsto a_i$, $e_P$ evaluates elements of $A$ at  $P$:
$$ e_P\colon g(x_1,\cdots,x_n)\mapsto g(a_1,\cdots, a_n)\in L.$$
The fact that $P\in V(L)$ means that $I$ is in the kernel $\tilde P$ of $e_P$, if one is thinking, a priori, of $P \in L^n$.
The quotient $A/\tilde P$ is isomorphic to the image $\tilde A$ of $e_P$, the sub-ring $\tilde A$ of $L/K$ generated by the $a_i$ (the coordinates of $P$). Now – if $L$ is an algebraic field extension, this image ring $\tilde A$ is in fact a field, so that $A/\tilde P$ is a field (the 'residue field'), and $\tilde P$ is maximal. [I hope I am answering k-rational's question here.]
Conversely, if $\eta\colon A \to L$ is an $K$-algebra homomorphism, then, of course, $$\eta\colon x_i\mapsto a_i\in L.$$ Hence $\eta$ corresponds to point $P_\eta = (a_1,\cdots , a_n)\in L^n$.
On the other hand, the ideal $\tilde P$ does not necessarily quite correspond to a point in $V(L)$.
For instance, say $V = \mathbb A_{\mathbb Q}^1$, so that $A = \mathbb Q [x]$. Suppose $L = {\mathbb C}$. Say that $P = (i)\in V(L)$ [where $i$ is the 'chosen' $i$ of complex analysis]. Then $e_P \colon A\to L$, defined by $e_P\colon x\mapsto  i$, has kernel $(x^2+1)$, and $A/\tilde P\simeq \mathbb Q (i) \subset L$.  But, of course, $ x \mapsto -i$ would have been 'another' point giving rise to the same ideal.  Over an algebraically closed field $K$, maximal ideals of $A$ do correspond to points of $V(K)$ - this is courtesy of the Nullstellensatz.
The correspondence $$\text{hom}_{K-\text{alg}}(A,L)\longleftrightarrow V(L)\text{, the $L$-points of $V$,}$$
is pretty much the scheme, functorial, point of view/definition. One writes $V(L)$, or even $V(L)_K$, if one's being pedantic/careful, to denote 'either.'
[An example, where $L$ is (usually) far from a field, is if $L=A$, and $\eta\in V(A)$ is the identity map:
$$\eta: x\to x,$$
or $P=(x_1,\cdots, x_n)\in A^n$, and $e_P=\lambda$. This is 'useful', e.g., for some arguments in group schemes.]
As I said in the comment section, Silverman's  Arithemtic of Elliptic Curves uses the notation $K(P)$ to denote the field generated by the coordinates of $P$ (see Chapter 1, section 2; actually, $P$ is actually a point in projective space – so there is a further subtlety, but...) In any case, I, as someone with a number-theory background, feel the notation is pretty common.
A last comment: for me, 'residue field' implies that there is a local ring $R$, with maximal ideal $m$, and that $k(m)= R/m$ is the residue field. Here 'my' $A$ is not (usually) local. But since the $\tilde P$ is maximal, $ A/m \simeq A_{\tilde P}/ mA_{\tilde P}$, so all is well. Be that as it may, https://en.wikipedia.org/wiki/Residue_field seems OK with the more general usage - if $m$ is maximal. Likewise, suppose now that $A$ is an integral domain, and $L$ a transcendental extension of the quotient field of $A$. I am pretty sure that people who used to talk about 'generic points' $P\in V(L)$, where $P$ is transcendental over $A$, probably where perfectly happy to write $K(P)$ for the field generated by the 'coordinates' of $P$. It seems reasonable to me - although I have no reference at hand, I bet one could find that in Lang or Weil, or perhaps even the Italians - but I don't know. See https://en.wikipedia.org/wiki/Generic_point.
