Changing basis of indeterminates in polynomial ring

2018-02-17 21:10:44

In the polynomial ring $R={\mathbb Q}[a,b,c,d,\beta_0,\beta_1,\beta_2,\beta_3]$, consider

the quotient commutative algebra $A=R/I$ where $I$ is the ideal generated by the four polynomials

$$

\begin{array}{lcl}

P_0 &=& \frac{1}{(ac)^2B_1}\big(\beta_0 M-(a^2c^2B_1)^4\big) \\

P_1 &=& \frac{1}{c^2C_1D_1}\big(\beta_1 M-(c^2C_1D_1)^4\big) \\

P_2 &=& \frac{1}{a^2C_2D_1}\big(\beta_2 M-(a^2C_2D_1)^4\big) \\

P_3 &=& \frac{1}{B_1C_1D_2} \big(\beta_3 M-(B_1C_1D_2)^4\big) \\

\end{array}

$$

where

$$

\begin{array}{ccc}

B_1 =\frac{ab+cd}{2}, & C_1 =\frac{ac+bd}{2}, & D_1 =\frac{ad+bc}{2}, \\

B_2 =\frac{ab-cd}{2}, & C_2 =\frac{ac-bd}{2}, & D_2 =\frac{ad-bc}{2}, \\

\end{array}

$$

and

$$M=(abcd)^2B_1C_1D_1B_2C_2D_2.$$

If we consider the field of fractions $F$ of $A$, then $F$ is obviously isomorphic to ${\mathbb Q}(a,b,c,d)$, since in $F$ each of the $\beta_k$ is a rational function in $a,b,c,d$. But things become very different when we consider $F$ as an extension of ${\mathbb