[Math] Motivation behind the definition of flat module

commutative-algebraflatnessmodulesmotivation

Can someone explain what is the motivation behind the definition of a flat module? I saw the definition but I don't really know why it is important to work with these structures.

Best Answer

Flatness in commutative algebra satisfies a geometric condition: the fibers of a morphism between two varieties (schemes) don't vary too wildly. A flat morphism $f \colon X \to Y$ of varieties (schemes) can be thought as a continuous family of varieties (schemes) $\{ f^{-1} (y) \}_{y \in Y}$. An important theorem says that if $f \colon X \to Y$ is a flat morphism between two irreducible varieties then its fibers have dimensions equal to $\dim X - \dim Y$.

For example, fix an algebraically closed field $k$, and consider the morphism $f \colon \mathbb{A}^1 \to \mathbb{A}^1$ defined by $x \mapsto x^2$. It corresponds to the ring homomorphism $k[x^2] \hookrightarrow k[x]$. You should be able to prove that $k[x]$ is a flat $k[x^2]$-algebra (it is also free), hence the morphism $f$ is flat. The fibers are almost always made up of two points.

Consider the morphism $g \colon \mathbb{A}^2 \to \mathbb{A}^2$ defined by $(x,y) \mapsto (x, xy)$ (it is an affine chart of the blowing up of the plane). It corresponds to the ring homomorphism $k[x,xy] \hookrightarrow k[x,y]$. The fiber of the point $(0,0)$ is the line $\{ x= 0 \}$ which has dimension $1$, while the fiber of the other points is empty or made up of a single point. You should be able to check that $k[x,y]$ is not flat as $k[x,xy]$-algebra.

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