Localization with maximal ideal

Question

Let R be local ring with maximal ideal P. Show that every element of R\P is invertible. Now let e be an element of R satisfying $e^2=e$. How can we prove $e \in {0,1}$.

Thanks to all for these helpful hints. – S786 Jan 12 '14 at 15:58 — S786, Jan 12 '14 at 15:58

Bill Dubuque · Answer 1 · 2014-01-12T16:31:01.383

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Hint $\,\ \color{#c00}{0 = e(1\!-\!e)}\in P\,\Rightarrow\, $ exactly $\,\overset{\large\quad\, {\rm else}\ \ 1-e + e\, =\, 1\in P}{{\rm one\ of}\ \,1\!-\!e,\, e\,\in\, P,\,}$ so other is a unit, so $\rm\color{#c00}{cancelling\ it}\ldots$

edited Jan 12 '14 at 16:31

answered Jan 12 '14 at 15:51

Bill Dubuque

272,048

score 0 · Answer 2 · answered Jan 12 '14 at 15:40

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Hint: Local rings have one maximal ideal and non-units belong to some maximal ideal.

answered Jan 12 '14 at 15:40

img_teacher

384

score 0 · Accepted Answer · answered Jan 12 '14 at 15:49

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another hint: Let $R$ be a commutative ring and let $J$ be its Jacobson radical. For any element $a\in J$ , $1-a$ is an invertible.

In you case, $R$ is local ring, so that maximal ideal $P$ is exactly the radical.

answered Jan 12 '14 at 15:49

Z. Ameen

56

score 0 · Answer 4 · answered Jan 12 '14 at 16:00

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Suppose by way of contradiction that $e\notin \{0,1\}$. Then $eR\oplus(1-e)R=R$ splits $R$ into two proper right ideals. But both halves are contained in $M$... how could they add up to $R$ when $M$ is closed under addition?

answered Jan 12 '14 at 16:00

rschwieb

153,510

Localization with maximal ideal

4 Answers4