2

Suppose that $R$ is a commutative ring and $I$ is a finitely generated ideal of $R$ such that $I^2=I$. Show that $I=Re$ for some idempotent element $e$.

I think I should use induction on the number of generators of $I$. I have proved that it's true for $n=1$:

Suppose $I=\langle a\rangle$ for some $a \in I$. Since $I^2=I$ we have $\langle a^2\rangle = \langle a\rangle$ and therefore $\exists r \in R: a=ra^2$. Multiplying both sides by $r$ and setting $e=ra$ we see that $e^2=e$ and $I=\langle e\rangle=\langle ra\rangle\subseteq \langle a\rangle =I$. Therefore, $I=Re$.

Now for $n=2$, things look difficult. Suppose $I=\langle a_1,a_2\rangle$.

Again, we have $\langle a_1^2,a_1a_2,a_2^2\rangle=\langle a_1,a_2\rangle$ therefore there exists some coefficients in $R$ such that we have:

$$a_1 = r_{11}a_1^2 + r_{12}a_1a_2 + r_{22}a_2^2$$ $$a_2 = s_{11}a_1^2 + s_{12}a_1a_2 + s_{22}a_2^2$$

After factorization:

$$a_1 = a_1(r_{11}a_1 + r_{12}a_2) + r_{22}a_2^2$$ $$a_2 = s_{11}a_1^2 + (s_{12}a_1 + s_{22}a_2)a_2$$

It isn't easy to see an idempotent element in these relations, let alone finding one that generates $I$.

user375190
  • 249
  • 1
  • 11
  • See http://math.stackexchange.com/questions/42727/finitely-generated-idempotent-ideals-are-principal-proof-without-using-nakayama/42735 and also http://math.stackexchange.com/questions/25045/problem-on-idempotent-finitely-generated-ideal (the answer, not the question). – darij grinberg Jan 13 '17 at 10:31

1 Answers1

1

You Can use the Nakayama lemma applied to the $R$-module $R$ and the ideal $I$, there exists $r\in R$ such that $r=1-i, i\in I$ and for every $j\in I$, $rj=(1-i)j=0$, this implies that $ji=j$, in particular $i^2=i$. This shows that $i\in I$ generate $I$ and $i^2=1$.

https://en.wikipedia.org/wiki/Nakayama's_lemma

jacobian
  • 156
  • 8
Tsemo Aristide
  • 87,475