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An element of ring $e$ is called idempotent iff $e^2=e$.

Let $R$ be a commutative ring that contains the identity element and a non-trivial idempotent element.

I want to show that each of its prime ideals contain also an idempotent element.


We have that $P$ is a prime ideal of $R$ iff $\forall a,b\in R$ : $a\cdot b\in P\Rightarrow a\in P \text{ or } b\in P$.

We have that $1\in R$ and $e^2=e\in R$.

So, $1\cdot e^2\in P \Rightarrow e^2\in P$, since $1\notin P$.

Since $e^2=e$, we also have that $e\in P$.

Does this imply that $e^2=e\in P$ ?

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    $\begingroup$ "So $\;1\cdot e^2\in P\,"...$ .... why ? $\endgroup$ – DonAntonio Apr 12 '16 at 17:29
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You can't assume $e^{2}$ $\in$ $P$. However $0$ $\in P$ and since $e^{2}-e=0$ $\in$ $P$, we have either $e$ $\in$ $P$ or $e-1$ $\in$ $P$. In either case we are done (why?).

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    $\begingroup$ You had me at $0\in P$ (since zero is "an idempotent"). However I suppose $e-1$ is a "nontrivial" idempotent if and only if $e$ is. $\endgroup$ – hardmath Apr 12 '16 at 17:43
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    $\begingroup$ @hardmath it's, $1-e$ is "non-trivial" idempotent if and only if $e$ is. That easily follows from just squaring the elements. $\endgroup$ – iron feliks Apr 12 '16 at 17:48
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    $\begingroup$ @Mary Star , your question assumes the existence of a non trivial idempotent element ,say $e$ in the ring. We can't assume $e \in P$. What we prove is that $P$ either contains.$e$ or $1-e$ both of which are idempotent , as you should check . $\endgroup$ – iron feliks Apr 12 '16 at 17:57
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    $\begingroup$ We have that $e^2=e$. Then $(1-e)^2=1-2e+e^2=1-2e+e=1-e$, so $1-e$ is idempotent, right? @ironfeliks $\endgroup$ – Mary Star Apr 12 '16 at 18:05
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    $\begingroup$ yes! @Mary star $\endgroup$ – iron feliks Apr 12 '16 at 18:06
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Hint: $e^2=e$ implies $e(e-1)=0$.

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No. In $1\cdot e²\in P$ you assumed that $e²=e\in P$, but this is the thesis.

Since $e²=e$, then $e²-e=e(e-1)=0$. Therefore $\overline{e(e-1)}=\overline{0}$ in $R/P$. If $P$ is prime, $R/P$ is a domain; so $\overline{e}=\overline{0}$ or $\overline{e-1}=\overline{0}$. If $\overline{e-1}=\overline{0}$, we have $e-1\in P$ and $$(1-e)²=(e-1)²=e²-2e+1=e-2e+1=1-e\in P.$$ If $e\in P$, done. We prove that every prime ideal in $R$ contain an idempotent element.

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