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How to compute the Krull dimension of the $ \mathbb{C} $ - algebra $ \mathbb{C} [X_1 , \dots , X_n ] / ( f_1 , \dots , f_m ) $ where $ m \leq n $, and such that the polynomials $ f_1 , \dots , f_m $ are irreducible ?

Thank you in advance for your help.

user26857
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Albert007
  • 367

2 Answers2

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To expand a little on Mariano's comment: the trick is to compute Gröbner bases, and this is best done on a computer.

Basically, a Gröbner basis is a nice set of generating functions for of your ideal. The really good thing about Gröbner bases, is that they tell you which terms in this generating set is "largest", and that by forgetting all other terms - thereby getting a monomial ideal (generated by the largest terms of the generators), with many of the same properties as your original ideal.

Specifically, it has the same dimension (and Hilbert polynomial) as your original ideal - and it is much easier to compute the dimension of a monomial ideal than to compute the dimension of an arbitrary ideal. This is how computer algebra systems such as Macaulay2 compute dimension.

For example, in user26857's answer, the leading terms generate the ideal $(y^2,x^2y,x^3)$ (under some ordering), which has radical $(y,x)$, which is of dimension $1$!

Fredrik Meyer
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$\dim\mathbb C[x,y,z]/(x,y,z)=0$ and $\dim\mathbb C[x,y,z]/(xz-y^2, yz-x^3, z^2-x^2y)=1$

This shows that one can't say much about it.

user26857
  • 52,094