Re: representing ideal as an intersection of prime ideals

On 28.10.2006 10:07, Jannick Asmus wrote:
On 27.10.2006 21:47, eugene wrote:

[snip]

Well, to show that R=k[x,y,z]/I (with a field k) is reduced note that
R'=k[x,z]/(z^3-x^5) is an integral domain and R is a free R'-module with
basis 1 and y.

[snip]

Hi again, i'm sorry for reviving old threads. But is think that there
still must be something wrong with you idea...

at first suppose (g_1 + g_2y)^n = 0 (*) . where g_1,g_2 \in R' and R
is considered as a free module over R' with respect to the basis {1,y}
as you said.
So, exapnding the expression we know that in R, y^2 = xz, so that the
only powers in (*) which contain y are y,y^3,y^5,... and last time i
wrote the wrong expression and in fact the coefficient in y is more
complicated, namely we have :

y*( ng_1^(n-1) * g_2 + C_n^3 g_1^(n-3)g_2^2 * xz + ... +) = 0, and
in fact at least for me it is not evident that it follows that g_1 =
g_2 = 0.

Is it the idea you actually meant? or maybe i did somethign wrong.

Thanks

Well, I must admit that it is not so easy as I thought. But we can save
the idea - without too lengthy calculations. Too bad that we need to do
some algebra at one single point.

So, here we go. In order to reduce nasty expressions let's try the trick
of the degree of nilpotence.

I am using the notations of R and R' as above. Let f = g1 + g2.y be a
non-zero nilpotent in R with g1,g2 in R'. If g2=0, then g1 is nilpotent
in R', hence g1=0, too, since R' is an integral domain. Contraction.

So we can assume that g2=/=0. Let n>0 be minimal such f^n = 0 (n is
called the degree of nilpotence of f). Then n>1 and f^(n-1) =/= 0. Some
algebra yields that f^(n-1) = h1 + g2.h2.y with h1,h2 in R'. [The only
thing we need here is that the coefficient of f^(n-1) wrt y is divisible
by g2!]. Then 0 = f^n = f^(n-1).f = h1.g1 + g2^2.h2 xz + g2.(h1+g1.h2).y
. Since {1,y} is a basis of R over R' and g2=/=0 we have h1 = - g1.h2
(*). This implies that both h1 and h2 are non-zero due to the choice of
n. Plug (*) into 0 = h1.g1 + g2^2.h2 xz and get g1^2 = xz g2^2. The
last relation yields f^m = a_m.g1^(m-1).f with some konstant a_m in k
for m of the form 3k, k>0.

Addition: a_m non-zero.
Correction: m odd (at least for infinitely many m's).

Thus infinitely many powers of f almost
reproduce f. Hence ...

HTH.

J.
.

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