Skip to main content
MathOverflow

Return to Answer

Post Undeleted by Vinteuil
deleted 78 characters in body
Source Link
Vinteuil
Vinteuil
  • 754
  • 3
  • 11

If $n=\text{dim}_{K}(H_1(\mathbf{x},A))$ then $\text{dim}_{K}(H_2(\mathbf{x},A))=\frac{n(n-1)}{2}$. This was proved by Assmus in 1958. In general, $H_*(\mathbf{x},A)$ is the exterior algebra over $H_1(\mathbf{x},A)$, even under weaker hypothesis (Corollary 3 in Blanco-Majadas-RodicioBlanco-Majadas-Rodicio).

If $n=\text{dim}_{K}(H_1(\mathbf{x},A))$ then $\text{dim}_{K}(H_2(\mathbf{x},A))=\frac{n(n-1)}{2}$. This was proved by Assmus in 1958. In general, $H_*(\mathbf{x},A)$ is the exterior algebra over $H_1(\mathbf{x},A)$, even under weaker hypothesis (Corollary 3 in Blanco-Majadas-Rodicio).

If $n=\text{dim}_{K}(H_1(\mathbf{x},A))$ then $\text{dim}_{K}(H_2(\mathbf{x},A))=\frac{n(n-1)}{2}$. This was proved by Assmus in 1958. In general, $H_*(\mathbf{x},A)$ is the exterior algebra over $H_1(\mathbf{x},A)$, even under weaker hypothesis (Corollary 3 in Blanco-Majadas-Rodicio).

Post Deleted by Vinteuil
Post Undeleted by Vinteuil
Post Deleted by Vinteuil
Source Link
Vinteuil
Vinteuil
  • 754
  • 3
  • 11

If $n=\text{dim}_{K}(H_1(\mathbf{x},A))$ then $\text{dim}_{K}(H_2(\mathbf{x},A))=\frac{n(n-1)}{2}$. This was proved by Assmus in 1958. In general, $H_*(\mathbf{x},A)$ is the exterior algebra over $H_1(\mathbf{x},A)$, even under weaker hypothesis (Corollary 3 in Blanco-Majadas-Rodicio).