Concerning algebraic dependence of two algebraic elements

Question

Let $A$ be a unique factorization domain (UFD), $b_1 \neq b_2$ two algebraic elements over $A$, with minimal polynomials $h_1(T_1)$ and $h_2(T_2)$. Assume that $\sum_{i=0}^{n} \sum_{j=0}^{m} a_{ij}b_1^ib_2^j=0$, where $a_{ij} \in A$.

Is it true that there exist $u,v \in A[T_1,T_2]$ such that $\sum_{i=0}^{n} \sum_{j=0}^{m} a_{ij}T_1^iT_2^j = uh_1(T_1)+vh_2(T_2)$? If not, what is a possible counterexample?

Remark: For example, if $m=0$, then $\sum_{i=0}^{n} a_{i0}b_1^i=0$, so $\sum_{i=0}^{n} a_{ij}T_1^i$ is a multiple of $h_1(T_1)$, and my above claim is true, with $u \in A[T_1]$ and $v=0$. However, my above claim may be false in case $n > 1$ and $m > 1$.

Thank you very much!

Answer 1

Take $A=\mathbb{C}[t]$, $b_1=t^{\frac{1}{2}}, b_2=t^{\frac{1}{4}}$. Then, $b_2^2-b_1=0$, $h_1(T_1)=T_1^2-t, h_2(T_2)=T_2^4-t$. But, $T_2^2-T_1\not\in (h_1(T_1),h_2(T_2))$.

Answer 2

Consider the following. Let $A=\Bbb{Z}$, $b_1=\sqrt2$ and $b_2=1+\sqrt2$, so $h_1(T_1)=T_1^2-2$ and $h_2(T_2)=(T_2-1)^2-2$. We have the obvious relation $$b_1-b_2+1=0,$$ so let's look at the polynomial $f(T_1,T_2)=T_1-T_2+1\in\Bbb{Z}[T_1,T_2]$.

I claim that $f$ is not in the ideal generated by $h_1$ and $h_2$. Assume contrariwise that $$ f(T_1,T_2)=u(T_1,T_2)h_1(T_1)+v(T_1,T_2)h_2(T_2)\qquad(*). $$ for some appropriate $u,v$. The catch is that in addition to $h_1(b_1)=0$ we also have $h_1(-b_1)=0$ because $\pm\sqrt2$ are algebraic conjugates. Therefore the right hand side of $(*)$ vanishes at the point $(T_1,T_2)=(-\sqrt2,\sqrt2+1)$. The left hand side obviously does not, so we have a contradiction.