Skip to main content
MathOverflow

Return to Answer

deleted 12 characters in body
Source Link
user87684
user87684

With the flatness hypothesis, it looks both statements are true.

The first being true implies the morphism

$$\text{Spec}(R[x_1,\dots, x_n]/(f,g)) \to \text{Spec}(R)$$

is a flat lci. But then the ideal $(f,g)\subset R[x_1,\dots, x_n]$ has to be regular, hence, calling $A := R[x_1,\dots, x_n]$, $B := A/(f)$, the Koszul complex of $(f,g)$ is computed by $K_{\bullet}(f)\otimes_B B/(g)$, whose first homology (which vanishes) is $\text{Tor}_1^{A}(A/(f), A/(g))$.

With the flatness hypothesis, it looks both statements are true.

The first being true implies the morphism

$$\text{Spec}(R[x_1,\dots, x_n]/(f,g)) \to \text{Spec}(R)$$

is a flat lci. But then the ideal $(f,g)\subset R[x_1,\dots, x_n]$ has to be regular, hence, calling $A := R[x_1,\dots, x_n]$, $B := A/(f)$, the Koszul complex of $(f,g)$ is computed by $K_{\bullet}(f)\otimes_B B/(g)$, whose first homology (which vanishes) is $\text{Tor}_1^{A}(A/(f), A/(g))$.

With the flatness hypothesis, it looks both statements are true.

The first being true implies the morphism

$$\text{Spec}(R[x_1,\dots, x_n]/(f,g)) \to \text{Spec}(R)$$

is a flat lci. But then the ideal $(f,g)\subset R[x_1,\dots, x_n]$ has to be regular, hence, calling $A := R[x_1,\dots, x_n]$, $B := A/(f)$, the Koszul complex of $(f,g)$ is $K_{\bullet}(f)\otimes_B B/(g)$, whose first homology (which vanishes) is $\text{Tor}_1^{A}(A/(f), A/(g))$.

Source Link
user87684
user87684

With the flatness hypothesis, it looks both statements are true.

The first being true implies the morphism

$$\text{Spec}(R[x_1,\dots, x_n]/(f,g)) \to \text{Spec}(R)$$

is a flat lci. But then the ideal $(f,g)\subset R[x_1,\dots, x_n]$ has to be regular, hence, calling $A := R[x_1,\dots, x_n]$, $B := A/(f)$, the Koszul complex of $(f,g)$ is computed by $K_{\bullet}(f)\otimes_B B/(g)$, whose first homology (which vanishes) is $\text{Tor}_1^{A}(A/(f), A/(g))$.