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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?

Nilotpal Chakraborty
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    Raising $1=as+bt$ to the power $2k-1$, we see that each term on the right involves either $s^k$ or $t^k$, so we can write $1=cs^k+dt^k$. – Andrew Hubery Feb 28 '24 at 17:59
  • I have also realised another way, for $As^{k_1} + At^{k_2} \neq A$, it must be contained in some maximal ideal $M$ then $s^{k_1} ,t^{k_2} \in M \implies s,t \in M \implies A=M$, contradiction. – Nilotpal Chakraborty Mar 01 '24 at 15:27

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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 $A^m+B^n=R$.

Asigan
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