A mild version of Bezout's identity in any commutative ring

Question

Let $A$ be any commutative ring (with $1$) and $x,y \in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b \in A$ such that $ax^k+by^l = 1$.

(Proof: Suppose otherwise. Then, $(x^,k,y^l) \subset \mathfrak p$ for some prime ideal $\mathfrak p$. But then this implies that $x,y \in \mathfrak p$ and therefore $1 = x+y \in \mathfrak p$, contradiction.)

My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?

Answer 1

Assuming $k,l$ non-negative integers.

Write $$1=(x+y)^{k+l}=\sum_{i=0}^{k+l}\binom{k+l}{i}x^iy^{k+l-i}$$

Now, for $i=0,\dots,k+l$ either $i\geq k$ or $k+l-i\geq l.$

So we just separate the terms. If we set:

$$\begin{align}a&=\sum_{i=k}^{k+l}\binom{k+l}{i}x^{i-k}y^{k+l-i}\\ b&=\sum_{i=0}^{k-1}\binom{k+l}{i}x^{i}y^{k-i}\end{align}$$

Then $ax^k+by^l=1.$

This actually works if $a_0x+b_0y=1$ for any $x,y,a_0,b_0\in R.$

For the case $x+y=1,$ you don't need the ring to be commutative, you just need $x,y$ to commute.

Answer 2

You have $(x+y)^{k+l}=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $\sum \binom{m}{r} x^ry^{m-r}=1$. Thus, you have $\sum_{r\leq k}\binom{m}{r} x^ry^{m-r}+\sum_{r>k} \binom{m}{r} x^ry^{m-r}=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.