Note, in the following, when I'm talking about solutions in multiple dimensions compared to ones in one dimension, what I mean is that each coordinate on its own should be considered a single-dimensional solution.
So for instance, if I get $\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} s_1\\ s_2 \end{bmatrix}$, then I consider $s_1$ and $s_2$ two separate solutions, eventhough they normally clearly are only one solution of a two-dimensional problem. This is so I can actually compare with one-dimensional solutions. The whole premise of the question wouldn't make sense if not for that.

The set of algebraic numbers is the set of roots of polynomials of arbitrary degree $n$ with integer coefficients $a_i$.

$$P_n{\left[x\right]} = \sum_{i = 0}^{n}{a_i x^i}\\ \mathbb{\overline{Q}_n} : \left\{ x | P_n{\left[x\right]} = 0 \right\}\\ \mathbb{\overline{Q}} : \text{values of } \mathbb{\overline{Q}_n}\text{ for any } n.$$

If you try plugging in algebraic numbers as coefficients for those polynomials, you won't find anything new. The solutions will all be algebraic numbers as well.

Is this also true if you allow higher-dimensional polynomials?

Say you have a polynomial of the form $a_{2 0} x^2 + a_{0 2} y^2 + a_{1 1} x y + a_{1 0} x + a_{0 1} y + a_{0 0} = 0$ where $a_{i j} \in \mathbb{Z}$ and $a_{2 0}, a_{0 2}, a_{1 1}$ can't all be zero.

If I'm not mistaken, this will not yet suffice since you'll get infinitely many solutions that form a $1$-dimensional subspace. So let's add a second polynomial of the same form to fix a couple points.

$$\begin{align*} a_{2 0} x^2 + a_{0 2} y^2 + a_{1 1} x y + a_{1 0} x + a_{0 1} y + a_{0 0} & = 0 \\ b_{2 0} x^2 + b_{0 2} y^2 + b_{1 1} x y + b_{1 0} x + b_{0 1} y + b_{0 0} & = 0 \\ a_{i j}, b_{i j} & \in \mathbb{Z} \end{align*}$$

and the six highest degree terms can't all be zero, so that at least one polynomial has degree $2$.

As long as these equations are independent, only a discrete finite set of solutions should be left over. I declare these solutions to be two-dimensional algebraic numbers of the second degree. Are these solutions the same as (one-dimensional) algebraic numbers?

By trying out this scheme for the first degree case (so just systems of linear equations), it is easy to see that the results will just be rational numbers which already are fully covered by (one-dimensional) first degree algebraic numbers (i.e. Integers and Rational Numbers). Based on that result, my hunch is that the answer to the title question will be negative: Algebraic numbers already fully cover this case for all degrees and all dimensions. But of course, the degree $1$ case is trivially represented in linear algebra, so it could easily be the case that non-algebraic solutions exist only for higher degrees. So is my hunch correct?

Bonus question 1: (Assuming my hunch is right) Will higher dimensional solutions change in degree? I.e. Can I represent (one-dimensional) algebraic numbers of degree $>n$ as solutions to systems of multidimensional polynomial equations of degree $n$?

Solution counting henceforth will be in the normal sense: $\begin{bmatrix} x\\y\end{bmatrix} = \begin{bmatrix}s_1\\s_2\end{bmatrix}$ is only one solution.

Bonus question 2: Up to how many solutions will a system of polynomial equations of degree $n$ in $d$ dimensions generally have?

  • 1
    $\begingroup$ In other words, will the (non-degenerate) intersection points between algebraic curves with integer coefficients have algebraic coordinates? $\endgroup$ – Henning Makholm Oct 1 '16 at 17:16
  • $\begingroup$ That sounds about right. $\endgroup$ – kram1032 Oct 1 '16 at 17:17

You won't get anything new. The points $(x,y)$ that are solutions of your pair of equations have coordinates that are algebraic numbers.

Given two (independent) polynomial equations $f(x,y)=0$ and $g(x,y)=0$ the resultant $R=R(f,g)$ is a polynomial with coefficients in $\overline{\Bbb{Q}}[y]$ with the property that the $y$-coordinate of any point in the intersection is a zero of $R$ (in general the coefficients of $R$ will be in the same field as those of $f$ and $g$). See the Wikipedia page on Resultants for links and more. Of course, when $y$ is algebraic $x$ will be also.

A different way of looking at it is that a point $(x_1,x_2,\ldots,x_n)$ with transcendental coordinates (over the field of definition $K$) will always be a generic point to a variety of dimension that is equal to the transcendence degree of the extension $K(x_1,x_2,\ldots,x_n)/K$. In other words, if you have transcendental coordinates in a solution, then the set of solutions is at least one-dimensional.

  • $\begingroup$ That was very well-worded! +1 Here's a rephrasing of what you said at the end: for a finite type $\mathbb{Q}$-scheme the closed points are the $\overline{\mathbb{Q}}$-points. So: $\overline{\mathbb{Q}}$-points only IFF only closed points IFF zero-dimensional. $\endgroup$ – Alex Youcis Oct 2 '16 at 9:39

Here is a way of phrasing what Jyrki said slightly differently (although I like his wording better--this is just to add a different perspective).

Let $R=k[x,y]/(f,g)$ and $X=\text{Spec}(R)$. Assume that $f$ and $g$ have no irreducible factors in common. I claim that this implies $X$ is $0$-dimensional. Indeed, let $Y\subseteq X$ be an irreducible component of $X$. Then, $Y$ corresponds to a minimal prime $\mathfrak{p}\supseteq (f,g)$. If $\mathfrak{p}$ weren't maximal then $\mathfrak{p}=(h)$ for some irreducible polynomial $h$ and this implies then that $h\mid f,g$ contradicting our assumptions. Thus, $\mathfrak{p}$ is maximal and thus $Y=\text{Spec}(k[x,y]/\mathfrak{p})$ is $0$-dimensional.

This then implies that $R$ is a $0$-dimensional $\mathbb{Q}$-algebra. This is somewhat obvious if one uses Jyrki's result relating Krull dimension and transcendence degree, but there's an easier method in this case (using just basic commutative algebra). Namely, since $R$ is $0$-dimensional and Noetherian it's Artinian. But, this then means that $R=A_1\times\cdots\times A_n$ with $A_i$ Artin local rings. Now, if $\mathfrak{m}_i$ denotes the maximal ideal of $A_i$ then $A_i/\mathfrak{m}_i$ is finite-dimensional over $\mathbb{Q}$ (this follows since $A_i$ is finitely generated over $\mathbb{Q}$--this is the Nullstellensatz). Note then we have the natural filtration of $A_i$ by $\mathbb{Q}$-subspaces


we will then be done (showing that $A_i$ is finite dimensional) if we can show that each $\mathbb{Q}$-space $\mathfrak{m}_i^\ell/\mathfrak{m}_k^{\ell+1}$ is finite-dimensional. But, note that it's dimension over $A_i/\mathfrak{m}$ (which is a finite-dimensional $\mathbb{Q}$-algebra) is finite since it's dimension (by Nakayama's lemma) is a minimal generating set for $\mathfrak{m}_i^\ell$ (which is finite since $A_i$ is Noetherian). Thus, the $A_i$, and thus $R$ is finite-dimensional over $\mathbb{Q}$.

So, now, suppose that $(x,y)\in K^2$ is a solution to $f=g=0$ (where $K$ is any field of characteristic $0$--perhaps $K=\mathbb{C}$ is the obvious choice from your setup). Then, note that $\mathbb{Q}[x,y]\subseteq K$ has the property that it's a quotient of $R$ and thus, finite-dimensional as a $\mathbb{Q}$-algebra. This implies that

$$\mathbb{Q}[x,y]\subseteq\{\alpha\in K:\alpha\text{ algebraic over }\mathbb{Q}\}$$

or, said differently, that $x,y\in\overline{\mathbb{Q}}$.

A perhaps nice way of thinking about this conceptually (with a lot swept under the rug) is the following. For a finite-type $\mathbb{Q}$-scheme the closed points of $X$ correspond (essentially--one must mod out by a Galois action) to the $\overline{\mathbb{Q}}$-points $X(\overline{\mathbb{Q}})$. So, one should see that $X$ produces non-algebraic points precisely when $X$ has non-closed points. That said, a (finite type $\mathbb{Q}$-scheme) scheme $X$ is zero-dimensional if and only if it consists of finitely many closed points. So, $X$ will have non-algebraic points if and only if it's zero-dimensional.

If you like algebra, then one can phrase this last condition as follows. The system $f_1=\cdots=f_k=0$ (with $f_i\in \mathbb{Q}[x_1,\ldots,x_n]$) will have non-algebraic points if and only if $\sqrt{(f_1,\ldots,f_k)}$ is NOT a product of maximal ideals. So, in your specific example, if $(a_i,b_i)$ denote the finitely many solutions of $f=g=0$ then

$$\sqrt{(f,g)}=\prod_i (x-a_i,y-b_i)$$

which shows why your example produces no non-algebraic numbers.

Just as a silly example: the non-algebraic point $(\pi,\pi^{-1})$ of $\mathbb{G}_{m,\mathbb{Q}}=\text{Spec}(\mathbb{Q}[x,y]/(xy-1))$ gives rise to the point $(0)=(xy-1)$ (since this is the kernel of the map $\mathbb{Q}[x,y]\to\mathbb{Q}[\pi,\pi^{-1}]$ with $x\mapsto \pi, y\mapsto \pi^{-1}$) which is a non-closed point.

EDIT: Here is another criterion that I think you might like. I claim that if $f_i\in k[x_1,\ldots,x_n]$ then $f_1=\cdots=f_k=0$ has only algebraic solution (in some field extension $K/\mathbb{Q}$) if and only if for one (equivalently all) algebraically closed extensions $L/\mathbb{Q}$ the system $f_1=\cdots=f_k=0$ has finitely many roots.

Indeed, label the implications

  1. $f_1=\cdots=f_k=0$ has only algebraic solutions.
  2. $f_1=\cdots=f_k=0$ has finitely many solutions in $\overline{\mathbb{Q}}$.
  3. $f_1=\cdots=f_k=0$ has finitely many solutions in any $K/\mathbb{Q}$.
  4. $f_1=\cdots=f_k=0$ has finitely many solutions in some $L/\mathbb{Q}$ with $L=\overline{L}$.

-1. implies 2. since we showed that in this $X:=\text{Spec}(R)$ with $R:=k[x_1,\ldots,x_n]/(f_1,\ldots,f_k)$ is $0$-dimensional and so discrete and so (by Noetherianess) finite.

-2. implies 1. follows from Noether normalization. Namely, if 1) didn't hold then $X$ would be positive dimensional, so surject onto some $\mathbb{A}^r_\mathbb{Q}$ with $r\geqslant 1$ and thus, consequently, have infinitely many $\overline{\mathbb{Q}}$-points.

-2. implies 3. because since $X$ is a finite collection of $\overline{\mathbb{Q}}$-points by the equivalence of 1. and 2.

-3. implies 4. obviously.

-4. implies 2. obviously.

So, your process of adding more equations to cut down the number of solutions to a finite amount doomed your quest for non-algebraic points from the beginning.

  • $\begingroup$ I see you are very enthusiastic about this topic. Unfortunately my grasp of this isn't nearly deep enough to understand like half of what you said there. I mean... I do know some of those words. :o) $\endgroup$ – kram1032 Oct 2 '16 at 11:13
  • $\begingroup$ @kram1032 Yes, I think I was struck with some sort of mania a few hours ago :P I don't know why I chose to put so much effort into this. Anyways, hopefully some of it is helpful--perhaps the last characterization which basically said that your goal was doomed from the start. $\endgroup$ – Alex Youcis Oct 2 '16 at 11:53

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