$As + At = A$ $\implies$ $as^k+bt^k=1$

Question

So I am studying this proof of Quillen's theorem. (page no. 27)

Here they say for a commutative ring $A$ we are given $As+At=1$.

So we get $as+bt=1$ for some $a,b\in A$.

But, in the proof they used for $k\in \mathbb{N}$ and $as^k+bt^k=1$ for some $a,b\in A$

Can anyone point what I am missing out?

Answer 1

Andrew's comment is a nice proof. Here is another one.

More generally, if $R$ is a commutative ring with $1$, and $A, B$ are ideals s.t. $A+B=R$, then $A^m+B^n=R$ for all $m,n\in\mathbb{Z}_+$. (Then the original question comes from applying this conclusion to $Aa$, $Ab$, $m=n=k$.) To see this, we first prove $A+B^n=R$ or all $n\in\mathbb{Z}_+$. This is because \begin{equation*} R=R^n=(A+B)^n\subseteqq A+B^n. \end{equation*} (To see why $\subseteqq$ holds, note that an element in $(A+B)^n$ must be a finite sum of terms like $(a_1+b_1)\cdots(a_n+b_n)\in b_1b_2\cdots b_n+A\subseteq B^n+A$.)

Since $A+B^n=R$ provided $A+B=R$, the substitution $B^n\leftarrow A$, $B\leftarrow A$, $n\leftarrow m$ implies $A^m+B^n=R$.