Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?

Question

Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?

I already did substitution for $X$ so we get the ring $R[x,x^{1/2},x^6,x^2]$ but I don't know to which ring this is isomorphic.

Answer 1

You are basically done. $R[x,x^{1/2},x^2,x^6]$ is just $R[x^{1/2}]$, which is just $R[\tilde{x}]$ (polynomials in one variable with coefficients in $R$).

Answer 2

You have $X=Y^2$, $T=Y^4$ and $Z=T^3=Y^{12}$, so you get $$ R[y^2,y,y^{12},y^4]=R[y] $$

Answer 3

Note that $T^3 - Y^{12}$ divisible by $T- Y^4$, so we have the equality of ideals $$\langle T- Y^4, T^3 - Z\rangle = \langle T- Y^4, T^3 - Y^{12}, T^3 - Z\rangle =\langle T- Y^4, Z- Y^{12}\rangle$$ and so $$\langle X-Y^2, T- Y^4, T^3 - Z\rangle = \langle X-Y^2, T- Y^4, Z- Y^{12}\rangle$$

Now the evaluation map

$R[Y,X,Z,T]\to R[Y]$, $X\mapsto Y^2$, $Z\mapsto Y^{12}$, $T\mapsto Y^4$ gives the isomorphism $$R[Y,X,Z,T]/\langle X-Y^2, T- Y^4, Z- Y^{12}\rangle \simeq R[Y]$$\