Properties of multivariate polynomials over $\mathbb Q$ modulo an ideal 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.
 A: 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.
A: 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.
