Given an $A$-algebra homomorphism $A[Y]\to A_a$ by sending $Y$ to $1/a$, where $a$ is an element of $A$. We want to find the kernel $I$.
The kernel $I$ is $(aY-1)$ . It is easy to see that the map $A[Y]/(aY-1)\to A_a\to A[Y]/(aY-1)$ is identity and the first map is surjective, hence the kernel is $(aY-1)$.
But when I try to directly figure out the kernel, it seems that it is not easy for me.
Following was what I tried. OKay, denote $B$ the ring $A[Y]/(aY-1)$. Now suppose $f(Y)=a_nY^n+a_{n-1}Y^{n-1}+\cdots+a_0\in I$, (stuck for some minutes), times $a^n$ to $f(Y)$, $a^n f(Y)=a_n+a_{n-1}a+\cdots+a_0 a^n \mod (aY-1)$. Since $a_n+a_{n-1}a+\cdots+a_0 a^n $ is zero in $A_a$, that means $a^{l+n}f(Y)\in (aY-1)$ for some integer $l>0$.
....... I fail.
I have on ideas to solve it in this way. Could anyone tell me what is the tricky here? Or give another proof here?
Edit: Again multiply $Y^{l+n}$ to $a^{l+n}f(Y)$, we obtain $f(Y)\in (aY-1)$.
It is really not easy.
Thanks.