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Is it true that a local ring, i.e., a commutative ring with a unique maximal ideal, doesn't contain idempotent elements $\neq 0, 1$ ? Why ?

Any hint ?

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7 Answers 7

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If $e$ is an idempotent which is not $0,1$, then $e(e-1)=e^2-e=0$ shows that both $e$ and $e-1$ are zero divisors and in particular not invertible. Hence they must be in the maximal ideal, but then $1=e-(e-1)$ is also in the maximal ideal - contradiction.

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  • $\begingroup$ Why do we necessarily need the zero divisors in the maximal ideal? $\endgroup$ Commented Dec 31, 2023 at 16:14
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    $\begingroup$ @Nothingspecial The ideal generated by a zero divisor is a nontrivial ideal (cannot contain the identity). Since there is a unique maximal ideal, then this zero divisor and its ideal are contained in the maximal ideal. $\endgroup$
    – Ofir
    Commented Jan 5 at 23:06
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Let $e$ be a nontrivial idempotent. Then $eR\oplus (1-e)R=R$ is a nontrivial splitting of $R$ into two proper ideals. But both $eR$ and $(1-e)R$ are contained in the maximal ideal: how could they add up to $R$? This shows no such $e$ can exist.

This works even for noncommutative local rings.

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  • $\begingroup$ I very much like this. However, is it even correct to state the equality in terms of a direct sum? I don't see an immediate reason to assume uniqueness of summands. Either way, a direct sum isn't needed - your argument would still hold. $\endgroup$ Commented Feb 15, 2016 at 23:02
  • $\begingroup$ @polynomial_donut It is indeed a direct sum. What cause do you have to doubt it? I do not understand your comment about uniqueness of summands since there is nothing like that in the definition, nor does it occur often. $\endgroup$
    – rschwieb
    Commented Feb 16, 2016 at 0:10
  • $\begingroup$ Well, I don't know which definition you are talking about, but here en.wikipedia.org/wiki/… and in probably any book on algebra, such as for example Atiyah/McDonald's Commutative Algebra to name a standard one, you will discover that direct sums are understood as either unique sum representations or as isomorphic to cartesian products of rings/modules, from which that unique sum representation would follow. There are also generic sums of submodules/rings, where uniqueness is not required and one would use ordinary plus symbols... $\endgroup$ Commented Feb 17, 2016 at 5:50
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    $\begingroup$ @polynomial_donut Don't worry, I'u using the standard definition. It looks like you just misexpressed yourself earlier by talking about "uniqueness of summands". That is totally different from uniqueness of sum representations. It's elementary to see that $R=eR+(1-e)R$ and $eR\cap (1-e)R=\{0\}$, and that is the notion of direct sum I am working with which appears in all your books and links. $\endgroup$
    – rschwieb
    Commented Feb 17, 2016 at 10:40
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    $\begingroup$ @polynomial_donut If $ex=(1-e)y$, multiply on the left by $e$ to see $ex=0$. So the intersection is trivial. If you are talking about a direct sum decomposition $A\oplus B$, the summands are $A$ and $B$, not sums of pairs of elements. $\endgroup$
    – rschwieb
    Commented Feb 19, 2016 at 9:10
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Let $(R,\mathfrak m)$ be local ring. Suppose that $e \in R$ is such that $e^2 = e$. If $e$ is a unit, then $e = 1$. If $e$ is not a unit, then $e \in m$ and by idempotency $Re = e (Re)$. Hence $Re = \mathfrak m (Re)$ and by Nakayama $Re = 0$ which implies $e = 0$.

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For every idempotent $e\notin \{0,1\} $ , $e\in\mathrm{Jac}(R) $ so $1-e$ is invertible. Then from $e(1-e)=0$ we conclude that $e=0$, a contradiction.

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  • $\begingroup$ Welcome to math.stackexchange. You may want to give some idea of what the Jacobian is, and why being in the Jacobian implies that $1-e$ is invertible. $\endgroup$
    – Joe Tait
    Commented Jul 3, 2014 at 12:53
  • $\begingroup$ Jacobson radical of$R$ denoted by $Jac(R)$. there is atheorem in commutative and noncommutative ring theory. $\endgroup$
    – sajad
    Commented Jul 5, 2014 at 21:48
  • $\begingroup$ why is $e \in Jac(R)$, can't we have an element that is outside? For instance, arbitrary element of a ring? $\endgroup$ Commented Jun 19 at 7:46
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Recall that there is also an elementary characterization of local rings, not needing the concept of (maximal) ideals: A ring $A$ is a local ring if and only if $1 \neq 0$ in $A$ and, for any $x, y \in A$ such that $x+y$ is invertible in $A$, $x$ or $y$ is invertible in $A$ (inclusive or). (If you don't fear the empty set, you can phrase this more succinctly as: A ring is a local ring if and only if, for any finite sum which happens to be invertible, at least one summand is invertible.)

With this characterization, the proof of your statement is easy: Assume $e^2 = e$. Then $e (1-e) = 0$. Since $e + (1-e)$ is invertible, $e$ is invertible or $1-e$ is invertible. In the first case, it follows that $1-e = 0$. In the second case, we have $e = 0$.

(The elementary characterization is crucial in constructive mathematics, where maximal ideals don't behave as well as in classical mathematics.)

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Remember that in a local ring $R$ the complement of the maximal ideal is the set of invertible elements, $R^{*}=R-m$.

The definition of maximal ideal implies the inclusion of $R^{*}\subset R-m$. If $x\in R-R^{*}$ then $(x)$ is different from $R$ and since $R$ has a unique maximal ideal $m$ we have $x\in (x) \subset m$. Therefore $R-R^{*}\subset m$. Thus, $R-m \subset R^{*}$.

Using this fact take $e$ an idempotent element, then $e^2=e$. Hence $e(e-1)=0 \in m$ implies that $e\in m$ or $e-1\in m$.

  • If $e\in m$ then $e-1\in R-m=R^{*}$ and $e=0$.
  • If $e-1\in m$ then $e\in R-m=R^{*}$ and $e=1$.
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  • $\begingroup$ In fact, $R\simeq k$. Moreover, $\frac{x}{a+bx}=\frac 01$ since $(x-1)x=0$. $\endgroup$
    – user26857
    Commented Dec 7, 2019 at 10:46
  • $\begingroup$ To conclude, your answer is wrong. And as a general suggestion, please trust the other answerers when five of them say the contrary. $\endgroup$
    – user26857
    Commented Dec 7, 2019 at 10:52
  • $\begingroup$ Excuse me, I think I rush into the answer. Thank you for your suggestion. $\endgroup$ Commented Jan 20, 2020 at 4:55
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Let $m$ be the unique maximal ideal of a local ring $A$. The Jacobson radical ($J(R)$) is defined as the intersection of all maximal ideals of $A$. Since $J(R)$ contains the nilradical $R$, which is itself an ideal, $J(R) = R = m$ by uniqueness.

Now, $x^2 = x \implies x(1- x) = 0$;

If $x \in R$, we know that $1-x$ is a unit and therefore cannot be a zero-divisor.

If $x \in A - m$, then $x$ is a unit per the definition of a local ring. The same conclusion follows.

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