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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.

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    $\begingroup$ Just to undersand your question: say $f(x,y)= x^2-2$, and $g(x,y) = y^2 -3$, with $K=\mathbb Q$. In this case, of course, the algebraic extension of $K$ should be "$\mathbb Q(\sqrt 2, \sqrt 3)$." Is this what you have in mind? $\endgroup$
    – peter a g
    Jul 15, 2021 at 13:09
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    $\begingroup$ Yes, that's what one [usually/always?] means by a field of a definition of a point $P$ in some (affine) space. $\endgroup$
    – peter a g
    Jul 15, 2021 at 13:56
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    $\begingroup$ I've seen both... But they should mean the same thing. (answering your question right before my 'deleted a comment' comment) $\endgroup$
    – peter a g
    Jul 15, 2021 at 14:05
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    $\begingroup$ Namely, if $P$ is a point in some (affine) space, the field $K(P)$ denotes the/a field extension of $K$ generated by the coordinates of $P$. $\endgroup$
    – peter a g
    Jul 15, 2021 at 14:09
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    $\begingroup$ @k-rational I've seen your question - I'll answer this weekend. $\endgroup$
    – peter a g
    Jul 16, 2021 at 14:13

1 Answer 1

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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.

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  • $\begingroup$ Thanks for your great effort. It is certainly praise-able. I will call it an answer to my as well. If you can just add some more information. Now it is clear to me that the coordinates of solutions of multivariable polynomials generate the field extension. The question is- When will such extension be Galois, in general ? For example, in my example the solutions of $x^2+y^2-2=0=x^2-2y^2-1$ generate the Galois extension $K(\sqrt 3,\sqrt 5)$ of $K$ with order of the Galois group $2$. what is the general criteria to check if the extension is Galois ? $\endgroup$
    – MAS
    Jul 19, 2021 at 16:05

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