Flatness over $K[\epsilon]/(\epsilon^2)$

Question

I'm reading this proof in Eisenbud and Harris "The Geometry and Schemes", and I got stuck on something that should be simple, but I can't see it. Any help would be much appreciated.

Let $I \subset K[x_1,...,x_n]$ be an ideal, and let $\tilde{I} \subset K[x_1,...,x_n,\epsilon]/(\epsilon^2)$ be an ideal generated by $(f_1 + \epsilon g_1,...,f_k + \epsilon g_k)$, where $f_i,g_i \in K[x_1,...,x_n]$ and the elements $f_i$ generate $I$.

Now, the first claim they make I don't understand is that a $ K[\epsilon]/(\epsilon^2)$-module is flat if and only if it preserves the exactness of the exact sequence: $$ 0 \rightarrow (\epsilon) \rightarrow K[\epsilon]/(\epsilon^2) \rightarrow K \rightarrow 0 $$

This is a necessary condition, I get that, but why is it sufficient?

The second problem arises when they apply this to the $ K[\epsilon]/(\epsilon^2)$-module $K[x_1,...,x_n,\epsilon]/\tilde{I}$. In that case one needs to show that the map $$ (\epsilon) \otimes K[x_1,...,x_n,\epsilon]/\tilde{I} \rightarrow K[x_1,...,x_n,\epsilon]/\tilde{I} $$ is injective, which they say is true if and only if for any $f \in K[x_1,...,x_n]$ we have: $$ \epsilon \cdot f \in \tilde{I}\implies f \in I $$ but I don't see why that would be.

For completeness this is part of the proof of Theorem VI-29. Thanks!

Answer 1

Actually your first question is answered by the fact, that you can test flatness against ideals, namely we have the following theorem:

Let $A$ be a ring and $M$ an $A$-module. $M$ is flat over $A$ if and only if for every finitely generated ideal $I \subset A$, the induced map $I \otimes_A M \to A \otimes_A M$ is injective.

You can find this theorem in Matsumura's book 'Commutative Ring Theory' as Theorem 7.7.

This answers your first question, since $(\varepsilon)$ is the only proper ideal of $K[\varepsilon]/(\varepsilon^2)$.

The second question comes down to show that $$f \in I \Longleftrightarrow \varepsilon \otimes f = 0$$ holds for $f \in K[x_1, \dotsc, x_n]$.

Answer 2

This might be helpful, but it might not.

Fact: If $B$ is an $A$-module, say by a homomorphism $f:A\to B$, then the following are equivalent:

$(1)$ $M'\to M\to M''$ is an exact sequence of $A$-modules if and only if $M'\otimes_A B\to M\otimes_A B\to M''\otimes_A B$ is exact

$(2)$ $B$ is flat over $A$ and $f^{-1}:Spec(B)\to Spec(A)$ is surjective

In general, an $A$-module $B$ satisfying condition $(1)$ is called faithfully flat. You may read about this more here.

In the case being considered here in particular, if we let $A=k[\epsilon]/(\epsilon^2)$, and we are considering an $A$-algebra, call it $B$, then $(1)$ is equivalent to $B$ being flat over $A$ because $Spec(A)$ is a single point, so $f^{-1}:Spec(B)\to Spec(A)$ is always surjective.