Given a commutative ring $R$ over an algebraically closed field $k$, an ideal $I$ is of colength $n$ if $\operatorname{dim}_k(R/I)=n$. Then it’s said when $R=k[[x,y]]$, every ideal of colength $n$ in $R$ contains $(x,y)^n$. I am confused here. Just take $n=2$, $(x,y)^2$ contains itself, but $k[[x,y]]/(x,y)^2$ has $1,x,y$ as a basis, shouldn't it be of colength $3$? Hope someone could help. Thanks!
Asked
Active
Viewed 86 times
0
-
1Since $(x,y)^3 \subset (x,y)^2$, this does not present a condraticion, no? – qualcuno Mar 16 '20 at 04:09
-
Could you explain it more clearly? – Yuyi Zhang Mar 16 '20 at 08:38
-
The result you quote says that if $I$ has colength $n$, then $(x,y)^n \subset I$. Your argument shows that $(x,y)^2$ has colength $3$, hence it should contain $(x,y)^3$. And that happens, since $(x,y)^3 \subset (x,y)^2$. – qualcuno Mar 16 '20 at 08:46
-
Right. Thank you! – Yuyi Zhang Mar 16 '20 at 11:58