Let $k$ be a field of characteristic zero, $R$ a commutative $k$-algebra, $a,b,c,d \in R$ such that the ideal generated by $a$ and $b$, $I_{a,b}=(a,b)$, equals the ideal generated by $c$ and $d$, $I_{c,d}=(c,d)$.
Is there an interesting relation between the $k$-subalgebra generated by $a$ and $b$, $k[a,b]$, and the $k$-subalgebra generated by $c$ abd $d$, $k[c,d]$?
Notice that if $u \in k[a,b]$, then it is of the form $u=\sum_{i\geq 1 \cup j \geq 1} u_{ij}a^ib^j+ u_{00}= \hat{u}+u_{00}$, and if $v \in k[c,d]$, then it is of the form $v=\sum_{i\geq 1 \cup j \geq 1} v_{ij}c^id^j+ v_{00}=\hat{v}+v_{00}$, where $\hat{u} \in I_{a,b}$, $\hat{v} \in I_{c,d}$, $u_{00},v_{00} \in k$.
Then for every $\lambda,\mu \in k$, we have: $\lambda u + \mu v = \lambda \hat{u} + \mu \hat{v} + \lambda u_{00} +\mu v_{00} = w + \nu$, where $w \in I_{a,b}=I_{c,d}$, $\nu \in k$.
Does this show that $k[a,b,c,d]=k[a,b]$ and $k[a,b,c,d]=k[c,d]$ so they are equal?
I don't think there is, you may always take $b=d=1$ and the equality of the ideals becomes trivial. Even worse, you may take $R$ to be a field extension of $k$. But $a$ and $c$ may be very different in any case, e.g. $k=\mathbb{Q}$, $R=\mathbb{C}$, $a=i$ the imaginary unit and $c=w$ the usual primitive third root of unity. The resulting $k$-algebras are not even isomorphic as $k$-vector spaces, since they have different dimensions.