local ring and subtraction with 1 as unit

1. Proposition

for \(r \in R\) it follows that \(r\) or \(r - 1\) is a unit
for \(r \in R\) it follows that \(r\) or \(r + 1\) is a unit
\(R\) is a local ring

2. Proof

2.1. 1) \(\implies\) 2)

Let \(r' = r - 1\) then \(r' = r - 1\) or \(r' + 1 = r - 1 + 1 = r\) is a unit

2.2. 2) \(\implies\) 3)

Let \(\mathfrak{m}_1,\mathfrak{m}_2\) be different maximal ideals. then they are coprime (cf. maximal ideals are coprime) and thus by chinese remainder theorem about surjectivity there exists a surjection

\begin{align*} \pi: R \twoheadrightarrow& R/\mathfrak{m} \oplus R/\mathfrak{m}' \\ r \mapsto (r + \mathfrak{m}, r + \mathfrak{m}') \end{align*}

choose a preimage \(r\) of \((-1,0)\). Then it follows, that \(r\) is not invertible (cf. unit under ring-homomorphism) and furthermore \(r + 1\) is also not invertible, as

\begin{align*} \pi[r + 1] =& (r + \mathfrak{m}, r + \mathfrak{m}') + (1,1) \\ =& (-1,0) +(1,1) \\ =& (0,1) \end{align*}

contradiction

2.3. 3) \(\implies\) 1)

corollary of local ring and non-units as maximal ideal because for the maximal ideal \(\mathfrak{m}\) it holds that either \(r\) or \(r + 1\) are not inculded as

\begin{align*} (r + 1) - r =& 1 \end{align*}