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I am a computer scientist, not a mathematician, please forgive my imprecisions. I came across the following structure and I need to understand it better.

Let $R = \mathbb{Q}[X,Y,Z,W]$ be multivariate polynomials over $\mathbb{Q}$ and consider the following elements: $X^2 - \alpha$, $Y^2 - \beta$, $Z^2 - X - \alpha$, $W^2 + X - \alpha$. I understand these generate an ideal $I$, with respect to which it is possible to define an equivalence relation $a \sim b \Leftrightarrow a-b \in I$. I need to operate with the classes $[a] = \{ b \in R : a \sim b \}$ defined by this equivalence, let me call the set of these classes $K$.

  • Is $K$ a field?
  • In case it is, is it a finite extension of $\mathbb Q$?
  • Can I find its dimension and a basis so as to treat it as a vector space?

I have some familiarity with similar issues involving finite fields and univariate polynomials but I do not know how to deal with the multivariate case. I did some quick research and I found about Groebner bases, do they have something in common with this problem?

edit

To make things clearer without reading the many comments below, the problem arised from trying to compute with $\sqrt{5}, \sqrt{2}, \sqrt{5\pm\sqrt{5}}$. It happened that the generators were chosen too naively and did not yield a field. A good choice is proposed in the answer below.

user9137
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2 Answers2

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My comments here are still incomplete, but it now appears that $R/I$ is never a field, because it has zero divisors: $(ZW)^2 - (2X)^2 = (ZW-2X)(ZW+2X)\in I$.

If the goal is to compute with $\sqrt{2}$, $\sqrt{5}$, and $\sqrt{5\pm \sqrt{5}}$, then the ring that should be used is $\mathbb{Q}[\sqrt{5+\sqrt{5}},\sqrt{2}] \cong \mathbb{Q}[X,Y]/(X^4-10X^2+20,Y^2-2)$, which is a field over $\mathbb{Q}$ of dimension $8$, with basis $\{X^i Y^j \mid 0\leq i \leq 3, 0\leq j \leq .1\}$

The important calculation in seeing that the above has everything you need is this one: $\sqrt{5+\sqrt{5}} \cdot \sqrt{5-\sqrt{5}} = 2\sqrt{5}$. Any field containing $\sqrt{5+\sqrt{5}}$ will also contain $\sqrt{5}$, therefore it will contain $\sqrt{5-\sqrt{5}}$.

It's not necessarily a field, depending on $\alpha$ and $\beta$. For example, if $\alpha = c^2$, then $\overline{(X-c)}$ is a zero-divisor in $R/I$.

It may also have nilpotents. If $\alpha^2 = \alpha$, then $(\overline{ZW})^2 = \overline{(\alpha-X)}\overline{(\alpha+X)} = \overline{(\alpha^2 - X^2)} = 0$.

It is, however, a vector space over $\mathbb{Q}$ with dimension $16$, with basis $\{\overline{X^a Y^b Z^c W^d} \mid a,b,c,d\in\{0,1\}\}$. I would work directly with this basis when doing computations, as the particular form of your ideal makes it very easy to reduce any polynomial to one written in terms of this basis (for this reason, you probably have little need for Gröbner bases).

I would speculate that $R/I$ is a field for most choices of $\alpha,\beta$, but I don't see how to quickly prove this.

Andrew Dudzik
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  • If $\alpha = c^2$ for some $c$ then I think $X - c$ and $X + c$ are zero divisors, and likewise with $Y$ and $\beta$, right? So there are quite a few $\alpha$ and $\beta$ that don't work, although I don't necessarily disagree that "most" still will. – Ben Millwood Oct 08 '14 at 23:02
  • Thanks! I will take some time to digest your answers and see what happens. By the way, what do you mean by the overbar? My first attempt was something like what you proposed, considering the exponents in ${0,1}$, but I noticed that I would have a redundant representation (different polynomials being actually equivalent and not noticing it in the algorithms) and I did not know well how to take full care of it, so I thought that I need to find a better basis. I also reduced the dimension to 12 by noticing that $ZW - 2X \in I$. Actually, $\alpha = 5$ and $\beta = 2$. – user9137 Oct 08 '14 at 23:30
  • @BenMillwood Yes, this is what I intended to say; I forgot briefly that $\beta$ was already in use. It should be the case that the set of "good" $\alpha,\beta$ in $\mathbb{A}^2$ contains a Zariski open... this is my intuition, anyway. – Andrew Dudzik Oct 08 '14 at 23:33
  • @user9137 I just used the overlines to distinguish between elements of $R$ and elements of $R/I$. How did you arrive at the conclusion that $ZW-2X \in I$? – Andrew Dudzik Oct 08 '14 at 23:38
  • You are right, it is not general. My bad, I see I derived it using my knowledge of $\alpha = 5$ and $\beta = 2$. Looking at the answer by Mariano I notice I can actually drop $X = \sqrt{\alpha}$ from the basis, right? I now suspect that this will give a unique representation for $R/I$. – user9137 Oct 09 '14 at 00:11
  • @user9137 How did you derive it even with those assumptions?—it seems like Mariano's answer supports that the dimension is 16. And I don't know what you mean by "drop", but then again I don't know your goals. – Andrew Dudzik Oct 09 '14 at 00:38
  • @Slade I am not sure I did not make any illegal assumption, but: $X=\sqrt\alpha$, $Z=\sqrt{\alpha+X}$, $W=\sqrt{\alpha-X}$, $ZW=\sqrt{\alpha(\alpha-1)}$, which is $2X$ when $\alpha=5$, therefore $ZW-2X=0 \in I$. If this holds, then also $5W+XW-2XZ=0$ and ${X^a Y^b Z^c W^d : a,b,c,d \in {0,1}}$ are not linearly independent. Where do I get it wrong? – user9137 Oct 09 '14 at 14:30
  • @user9137 As I said, I don't know your exact goals, but as stated this is wrong. All this shows, on the level of polynomials, is that $(ZW)^2 - (2X)^2 = (ZW-2X)(ZW+2X) \in I$, but not that either of the two factors are in $I$. I still think the dimension is $16$ for $\alpha=5$, $\beta=2$. Mariano's answer confirms that it is a power of $2$ for any $\alpha,\beta$, since you are making a series of four extensions by square roots, each of which has degree $1$ or $2$. – Andrew Dudzik Oct 09 '14 at 15:59
  • @user9137 In more "algebraic" terms, you have stated that $Z=\sqrt{\alpha+X}$, but if you are thinking of "solving" these equations, you should really write $Z=\pm\sqrt{\alpha+X}$. – Andrew Dudzik Oct 09 '14 at 16:02
  • Sorry, I shouldn't say that the dimension is a power of $2$. It is always 16, as far as I can tell, but if we add some extra structure (like passing to a prime ideal containing $I$) then we get a field whose dimension is a power of $2$. – Andrew Dudzik Oct 09 '14 at 16:14
  • Yes I think I understand what you are meaning, I arrived at similar arguments while reviewing my last comment. It seemed to me anyway that, say $A=ZW-2X,B=ZW+2X$, we have $AB \in I \Rightarrow AB \sim 0$, thus at least one of the factors should be null if $R/I$ is a field. My final goal is to compute symbolically with $\sqrt{5}, \sqrt{2}, \sqrt{5+\sqrt{5}}, \sqrt{5-\sqrt{5}}$, and I was trying to understand whether this can be done in a finite extension of $\mathbb{Q}$ and which basis could be used, mainly to be able to efficiently check whether a given element is rational. – user9137 Oct 09 '14 at 16:16
  • @user9137 I've just updated my answer. In short, your original ring can't possibly be a field, but there is a nice field that you can work with, requiring only two generators. – Andrew Dudzik Oct 09 '14 at 21:08
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This addressed part of your question. I'll write $a$ and $b$ instead of $\alpha$ and $\beta$.

If none of $a$, $b$ and $ab$ are squares in $\mathbb Q$,then the ring $\mathbb Q[X,Y]/(X^2-a,Y^2-b)$ is a field, namely $\mathbb Q[\sqrt a,\sqrt b]$. If over this field neither $a+\sqrt a$ nor $a-\sqrt a$ are squares then your ring is the field $\mathbb Q[\sqrt a,\sqrt b,\sqrt{a+\sqrt a},\sqrt{a-\sqrt a}]$, which is just $\mathbb Q[\sqrt b,\sqrt{a+\sqrt a},\sqrt{a-\sqrt a}]$.

If those conditions are not satisfied, then your ring is not a field.

Mariano Suárez-Álvarez
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