The unit groups of $A = \mathbb{Z}[S,T]/(S^2-3S, T^2 - 6T, ST - 3T)$

Question

Let $I = (S^2-3S, T^2 - 6T, ST - 3T)$, an ideal of $\mathbb{Z}[S,T]$, and $A = \mathbb{Z}[S,T]/I$. Decompose the unit group of $A[1/2]$ into a direct sum of cyclic groups.

I think the answer is $A[1/2]^* \cong \mathbb{Z}/2 \times \mathbb{Z}^3$. But I can't show it. Here is what I have tried:

First of all, $A$ is generated by $1,S,T$ as a$\mathbb{Z}$-module. (Actually this is a basis, but I don't need it in my proof.)
Define an injection $g: A \to \mathbb{Z}^3$ by $1 \mapsto (1,1,1), S \mapsto (0,3,3), T \mapsto (0,0,6)$.
Localizing it by $2 \in A$, we have an injection $A[1/2] \to \mathbb{Z}[1/2]^3$.

So $a+bS+cT \in A[1/2]$ is unit only if $a, a+3b, a+3b+6c \in \mathbb{Z}[1/2]^*$.
And computing $(a+bS+cT)(d+eS+fT)$ directly, we have that $a+bS+cT \in A[1/2]^*$ iff $a, a+3b, a+3b+6c \in \mathbb{Z}[1/2]^*$.

So we have an injective group homomorphism $ h: A[1/2]^* \to (\mathbb{Z}[1/2]^*) ^3$.
We have $ \left< (2,2,2), (-2,1,1), (1,-2,1), (1,1,-2) \right> \subseteq \operatorname{Im}h$.
If the converse is true, then using the isomorphism $\mathbb{Z}[1/2]^* \cong \mathbb{Z}/2 \times \mathbb{Z}: a \mapsto (\operatorname{sgn}a, \log_2 |a|)$, we have $A[1/2]^* \cong \left< (0,0,0,1,1,1), (1,0,0,1,0,0), (0,1,0,0,1,0), (0,0,1,0,0,1) \right> \subseteq \mathbb{Z}/2 ^3 \times \mathbb{Z}^3$.
Finally, this group is isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}^3$.

So my question is : $ \left< (2,2,2), (-2,1,1), (1,-2,1), (1,1,-2) \right> = \operatorname{Im}h$?

The previous part of the question also asks us to find all prime ideals of $A$ containing $2$. (The answer is $(2,S,T)$ and $(2,S-3,T)$, I think.)
So I doubt that this strategy is not good to determine $A[1/2]^*$...

Thank you very much!

Answer 1

Let $G=\left< (2,2,2), (-2,1,1), (1,-2,1), (1,1,-2) \right>$; we will show any element $(a,a+3b,a+3b+6c)$ in the image of $h$ must be in $G$. Writing $c=d/2-b/2$, we can instead write our element as $(a,a+3b,a+3d)$ Multiplying by a power of $(2,2,2)$ and possibly by $(-2,1,1)$, we may assume $a=1$. Then $a+3b$ is either a power of $2$ or negative a power of $2$. Multiplying by a power of $(1,-2,1)$, we can assume that $a+3b$ is either $1$ or $-1$. But since $a=1$, we must then have $a+3b=1$, since otherwise we would need to have $b=-2/3$ which is not in $\mathbb{Z}[1/2]$. Similarly, by multiplying by a power of $(1,1,-2)$ we may assume $a+3d=1$. Thus by multiplying by elements of $G$, we have reduced our element of the image of $h$ down to just $(1,1,1)$, and so every element in the image of $h$ must be in $G$.