Explicitly finding a quotient of two polynomial spaces

Let $p, p_1, p_2, p_3 \in \mathbb{Q}[x_1, x_2, x_3]$ (you can assume that $p_1,p_2,p_3$ are algebraically independent) and consider the space $\mathbb{Q}[p_1, p_2, p_3] \cdot p$. Now let $q_1, q_2, q_3 \in \mathbb{Q}[p_1, p_2, p_3]$ and $q \in \mathbb{Q}[p_1, p_2, p_3] \cdot p$ (again you can assume that they are algebraically independent). Then $\mathbb{Q}[q_1, q_2, q_3] \cdot q$ is a subspace of $\mathbb{Q}[p_1, p_2, p_3] \cdot p$ and we can look at their quotient: $$(\mathbb{Q}[p_1, p_2, p_3] \cdot p) \big/ (\mathbb{Q}[q_1, q_2, q_3] \cdot q)$$

Is there a general way to write this quotient space in a nicer form, say as $\mathbb{Q}[r_1, \ldots, r_s] \cdot r$ with some polynomials $r_1, \ldots, r_s \in \mathbb{Q}[p_1, p_2, p_3]$ and $r \in \mathbb{Q}[p_1, p_2, p_3] \cdot p$?

I'm currently looking at two specific cases of this problem, let me present (the easier) one of them here: $$p := (x_1^2 - x_2^2)x_3, \hspace{2mm} p_1 := x_1^2+x_2^2, \hspace{2mm} p_2 := x_1^2 x_2^2, \hspace{2mm} p_3 := x_3^2$$

\begin{array}{lll} q &:= (x_1^2 - x_2^2)x_3 &= p\\ q_1 &:= x_1^2+x_2^2+2x_3^2 &= p_1 + 2p_3\\ q_2 &:= -4x_1^2 x_2^2 x_3^2 + (x_1^2 + x_2^2)^2 x_3^2 &= -4p_2 p_3 +p_1^2 p_3\\ q_3 &:= x_1^2 x_2^2 - (x_1^2 + x_2^2)x_3^2 + x_3^4 &= p_2 - p_1 p_3 + p_3^2 \end{array}

Now in this case it's maybe helpful that $p = q$, but I'm not sure if the following holds due to that: $$R := (\mathbb{Q}[p_1, p_2, p_3] \cdot p) \big/ (\mathbb{Q}[q_1, q_2, q_3] \cdot q) \stackrel{?}{=} (\mathbb{Q}[p_1, p_2, p_3] \big/ \mathbb{Q}[q_1, q_2, q_3] ) \cdot p$$

Moreover, I found that $q_1, q_2, q_3$ are algebraically independent. And I calculated the dimension of the space $R$, which is: $$\dim(R) = \frac{1}{(1-t^2)^2} \frac{1}{1-t^6} t^5$$ This leads me to believe that it should be possible to write $R$ as $$R = \mathbb{Q}[r_1, r_2, r_3] \cdot r$$ where the degrees of the polynomials are $$\deg(r_1)=2, \hspace{2mm} \deg(r_2)=2, \hspace{2mm} \deg(r_3)=6, \hspace{2mm} \deg(r)=5.$$ Or is this last thought just wishful thinking?

Thanks in advance for any help!

• It would be nice you if could explain that "space" notation. – user26857 Feb 9 '17 at 7:41
• @user26857 Do you think I should have used the term "ring" instead? I think for the given problem one looks at things of the form $\mathbb{Q}[p_1, p_2, p_3] \cdot p$ as vector spaces over $\mathbb{Q}$. Or am I misunderstanding you? (If you're asking about the "$\cdot p$", that's just multiplication with the polynomial $p$) – Tom Bombadil Feb 9 '17 at 8:02
• I don't understand, which polynomials are algebraically independent? In your example you have $p^2-p_1^2p_3+4p_2p_3=0$, so $p,p_1,p_2,p_3$ are not algebraically independent. – san Feb 15 '17 at 14:03
• @san So sorry for this mistake! I mean just $p_1, p_2, p_3$ are algebraically independent. I edited the question just now. – Tom Bombadil Feb 15 '17 at 14:13