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Say that we have a polynomial ring $k[x_1,...,x_n]$ over a field k, and that J is some prime ideal of this ring, and call the quotient ring $S = k[x_1,...,x_n] / J$. Now, let K' be some prime ideal of this quotient ring. This corresponds to another ideal, let's call it K, containing J. How will V(K) (that is, the zero-set of the polynomials in the ideal of K) correspond to V(K')? Will they be equal? What can we say in general about this transition between the coordinate ring and the polynomial ring, in terms of zeros of polynomials?

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    I don't think you can define $V(K')$ as the zero set of polynomials, since $S$ isn't in general a polynomial ring... so how are you defining $V(K')$?2011-06-05
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    Zhen: Oh, you're completly right. I guess what I'm trying to say is to whether we can somehow relate the ideal here and "how it looks" in S, with how it looks in our polynomial ring. Say for example that in S, the ideal K' is principal. What does this mean for the ideal K? What will it be, and why so?2011-06-05
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    The Nullstellensatz is probably relevant here. For example, because $J$, $K$, $K'$ are all prime ideals, we could say that the dimension of $K'$ is equal to the dimension of $V(K) \cap V(J)$. You could probably work out something about whether $K'$ is principal when $K$ is by thinking about that sort of thing.2011-06-05

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