Re: Noetherian??
- From: quasi <quasi@xxxxxxxx>
- Date: Tue, 26 Jul 2005 16:44:56 -0700
On 26 Jul 2005 13:19:39 -0700, "crooner" <crooner_kwon@xxxxxxxxxxx>
wrote:
>I was trying to prove that if R is a commutative ring with identity
>which is Noetherian, then R+xR[x,y] is not a Noetherian ring.
>(R[x,y] ring of polynomial with two variables x,y.)
>
>First, I proved that xR[x,y] is an ideal of R+xR[x,y].
>Now, to prove not-Noetherianness, I thought I needed to
>prove xR[x,y] is NOT finitely generated. But I don't have any clue for
>that step. I guess it IS finitely generated by (x).
Let T be the ring R+xR[x,y] and let J be xR[x,y] so J is an ideal of
T. Does the ideal (x) in T contain the element xy? If so, show it. If
not then (x) is a proper ideal of (x,xy) which is a subideal of J,
where all ideals are viewed as ideals of T.
After that, look in turn at the elements xy^2, xy^3, ...
How does the ideal (x,xy,xy^2,xy^3,...) of T relate to J?
By the way, for this problem it doesn't matter whether or not R is
Noetherian, since there is no need to look at the ideals of R.
quasi
.
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