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Splitting in a complete local ring and decomposition its group of units

Abolfazl Tarizadeh

Abstract

Let $(R,M,k)$ be a complete local ring (not necessarily Noetherian). Then we reprove by a new method that the natural surjective ring map $R\rightarrow k$ admits a splitting if and only if $\Char(R)=\Char(k)$. In our proof there is no need for the existence of the coefficient field for equi-characteristic complete local rings, whose existence is the hardest part of the known proof. However, the main result of this article is that in the unequal characteristic case $\Char(R)\neq\Char(k)$, we prove that the natural surjective map between the groups of units $R^{\ast}\rightarrow k^{\ast}$ admits a splitting. As an application of the above theorem, we show that for any complete local ring $(R,M,k)$ the following short exact sequence of Abelian groups: $$\xymatrix{1\ar[r]&1+M\ar[r]& R^{\ast}\ar[r]&k^{\ast} \ar[r]&1}$$ is always split. Next, we show with an example that the above exact sequence does not split for many incomplete local rings.

Splitting in a complete local ring and decomposition its group of units

Abstract

Let be a complete local ring (not necessarily Noetherian). Then we reprove by a new method that the natural surjective ring map admits a splitting if and only if . In our proof there is no need for the existence of the coefficient field for equi-characteristic complete local rings, whose existence is the hardest part of the known proof. However, the main result of this article is that in the unequal characteristic case , we prove that the natural surjective map between the groups of units admits a splitting. As an application of the above theorem, we show that for any complete local ring the following short exact sequence of Abelian groups: is always split. Next, we show with an example that the above exact sequence does not split for many incomplete local rings.

Paper Structure

This paper contains 3 sections, 25 theorems, 6 equations.

Key Result

Theorem 1.1

Let $(R,M,k)$ be a complete local ring. Then the natural surjective ring map $R\rightarrow k$ admits a splitting if and only if $\operatorname{char}(R)=\operatorname{char}(k)$. Furthermore, if $\operatorname{char}(R)\neq\operatorname{char}(k)$, then the natural surjective map between the groups of u

Theorems & Definitions (56)

  • Theorem 1.1
  • Lemma 1.2
  • Lemma 1.3
  • Lemma 1.4
  • Corollary 1.5
  • Corollary 1.6
  • Lemma 2.1
  • proof
  • Lemma 2.2
  • proof
  • ...and 46 more