# relating flatness, equidimensional, and complete intersection

I am a bit confused and am trying to clarify some notions. First consider the following well-known statement.

A dominant map $f:X\rightarrow Y$ between regular varieties is flat if and only if it is equidimensional.

Question 1. Doesn't a regular variety mean that it is nonsingular?

Question 2. I am not sure what equidimensional means in this context. Does it mean that the fibers $f^{-1}(c)$ for each $c\in Y$ are equidimensional?

Question 3. What are some examples of a variety that is equidimensional but not a complete intersection?

Thank you.

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I believe regular and nonsingular are equivalent for perfect fields, but not in general. – Andrew Jul 5 '12 at 19:46
Thank you Andrew! – math-visitor Jul 5 '12 at 19:46

## 1 Answer

Googling, I find this article, which seems worth a look: www.emis.de/journals/UIAM/actamath/PDF/35-243-246.pdf. Particularly, saying that a morphism is equidimensional at a point of the domain means that nearby fibres of the morphism have the same dimension. This is different from equidimensionality for a scheme, which means that all irreducible components of the scheme should have the same dimension.

An example for question 3 is the twisted cubic, which is not a (global) complete intersection, but is an equidimensional variety of dimension $1$ (i.e. a curve).

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Thank you Andrew! I will definitely check out the article. – math-visitor Jul 5 '12 at 21:10