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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.

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If you do not demand that $a$ not be a zero divisor, the kernel may well be all of $A[Y]$. –  Mariano Suárez-Alvarez Mar 21 '12 at 17:06
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@MarianoSuárez-Alvarez It is okay for nilpotent element $a$, since $(aY-1)$ will become unit if $a$ is nilpotent. –  wxu Mar 21 '12 at 17:11

1 Answer 1

up vote 2 down vote accepted

The universal properties of polynomial algebras, quotient algebras and localizations imply that $\mathrm{Hom}(A[Y]/(aY-1),-)$ and $\mathrm{Hom}(A_a,-)$ are isomorphic functors. By the Yoneda lemma, this yields an isomorphism $A_a \cong A[Y]/(aY-1)$. As you can see, no calculation is needed at all. This is one of many explicit applications of the Yoneda lemma in algebra and related fields.

If you can't resist to make your hands dirty: Let $f = \sum_{i=0}^{n} a_i Y^i$ be in the kernel. Then we have $\sum_{i=0}^{n} a_i a^{n-i} = 0$ in $A_a$, which implies that for some $s \geq 0$ we have $\sum_{i=0}^{n} a_i a^{n+s-i} = 0$ in $A$. Assume for the moment that we may choose $s=0$. Then it follows

$f = \sum_{i=0}^{n-1} a_i Y^i + (- \sum_{i=0}^{n-1} a_i a^n ) Y^n = \sum_{i=0}^{n-1} a_i (1-(aY)^{n-i}) Y^i$.

Now each factor $1-(aY)^{n-i}$ is divisible by $1-aY$ (in general $q-1$ divides $q^n-1$, since $q^n-1=(q-1) \sum_{i=0}^{n-1} q^i$), thus also $f$ is divisible by $1-aY$.

For the case that $s$ is arbitrary, we are left to show the following: If $a^s f$ is divisible by $1-aY$, then the same is true for $f$. This is trivial if you work with quotient rings (in $A[Y]/(1-aY)$, we have that $a$ is invertible, thus regular), but let us continue this dull and unnecessary calculation: Write $a^s f = g (1-aY)$. Then $f = (1-(aY)^s) f + Y^s g (1-aY)$ and we have already remarked that $1-aY$ divides $1-(aY)^s$.

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