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I think I have a counter-example to the theorem

If $M$ is a maximal ideal in $R$ such that for all $x \in M$ , $x+1$ is a unit, then $R$ is a local ring with maximal ideal $M$. $R$ has a unique maximal ideal $M$.

A proof is supplied here, but I feel I have a counter-example. Can someone tell me why my counter-example is wrong?

Consider $R = \mathbb Z/6 \mathbb Z$. This has the ideal $M = \{0, 2, 4\}$ as a maximal ideal. $1 + M = \{1, 3, 5\}$ has every element a unit in $\mathbb Z / 6 \mathbb Z$. Thus, from the theorem, we are to conclude that $(M, \mathbb Z / 6 \mathbb Z)$ is a local ring.

But the ring $\mathbb Z / 6 \mathbb Z$ is not local! It has another maximal ideal $J = \{0, 3 \}$.

What am I missing?

Siddharth Bhat
  • 8,452

1 Answers1

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$3$ is not a unit of $\Bbb Z/6\Bbb Z$, as you've pointed out in the following paragraph.