Given a field $K$ we have the polynomial ring $K[x,y]$ in $2$ variables, which is also a left module (over itself). How can we prove that the ideal $(x,y)$ is not a free module?
Asked
Active
Viewed 3,232 times
12
-
Sorry. I meant the ideal <x,y> – Ben Berger Dec 31 '11 at 16:05
-
10Hint: any two elements are linearly dependent. So if it were free, it would be one-dimensional. – Chris Eagle Dec 31 '11 at 16:14
-
3...it would be of rank 1 – Mariano Suárez-Álvarez Dec 31 '11 at 16:22
-
1@Chris: Very nice! So, in a domain, an ideal is free iff it's principal. (Of course, Mariano is right, as usual...) --- Related answer (generalizing Chris's hint). – Pierre-Yves Gaillard Dec 31 '11 at 16:37
-
Thank you all very much! – Ben Berger Dec 31 '11 at 18:11
-
2Dear @Chris: I suggest that you upgrade your comment to an answer. – Pierre-Yves Gaillard Dec 31 '11 at 22:45
-
Related: https://math.stackexchange.com/questions/2255814 – Watson Nov 26 '18 at 13:18
1 Answers
13
This is a community wiki answer intended to remove this question from the unanswered queue.
As Chris Eagle hinted in the comments, free ideals in a commutative domain can only be generated by a single element.
If $a,b$ were two elements of a basis of a free ideal in a commutative domain, then $ba+(-a)b=0$ would be a nontrivial $R$-combination of the two, but that is absurd if they are members of an $R$-basis.
So, $(x,y)$, which isn't principal, cannot be a free ideal of $K[x,y]$.
rschwieb
- 153,510