Let $R$ be a commutative ring with $1 \ne 0$, and let $0 \ne e \in R$ be an idempotent element. Prove the following:

Question

Let $R$ be a commutative ring with $1 \ne 0$, and let $0 \ne e \in R$ be an idempotent element. Note that $eR=\{er|r \in R\}$ is also a commutative ring with identity element $e$.

(1) If I is an ideal of $R$, show that $eI=\{ex|x \in I\}$ is an ideal of the ring $eR$. Furthermore if $I$ is prime in $R$ and $e \not\in I$, show $eI$ is prime in $R$.

(2) If $J$ is an ideal of the ring $eR$, show that there exists a unique largest ideal $I$ of $R$ such that $eI=J$. Furthermore if $J$ is prime in $eR$, show that $I$ is prime in $R$.

End question

My thinking: For (1) If we have $er={er∣r \in R}$, then $e \in eR$ since $e=e^2$

Then, $(er)e=e^2r=er$ and $e(er)=e^2r=er$. Then $e$ is the identity of $eR$

From here, we have to prove two things: First, that $I$ is an ideal of $R$, therefore $eI$ is an ideal of $eR$. So, $I$ is a prime ideal and $e \not\in I$, which tells us that $ab \in I→a \in I$ or $b \in I$ And $eI$ is prime of $eR$, which tells us that we need to show $(er)(es) \in eI \to er \in eI or es \in eI$

Second: $J$ is an ideal of $eR$, so there exists a largest ideal $I$ of $R$ such that $eI=J$ (If $K$ is an ideal of $R$ and $eK=J$, then $K \le I$)

So, then we must show that if $J$ is prime in $eR$, then $I$ is prime in $R$

But I was also thinking of another way:

If we suppose $J$ is an ideal of $eR$, we need to construct an ideal $I$ of $R$ such that $J=eI$. If we define $I={r\in R/ er \in J}$, maybe we can use this to show $I$ is an ideal of $R$ and $J=eI$??

Answer 1

There is a straightforward way to construct $I$. Let $I$ be the set of all elements $x$ such that $ex\in J$. We show:

(1) We show that $I$ is an ideal. It is an additive subgroup because if $ex,ey\in J$, then $e(x+y)=ex+ey\in J$. If $ex\in J$ and $y\in R$, then $exy=(ex)(ey)\in J$ because $J$ is an ideal, hence $xy\in I$. Thus $I$ is an ideal.

(2) We show that $eI=J$. If $z=ey\in J$, then by definition $y\in I$. Also by definition, if $y\in I$, then $ey\in J$. Thus multiplication by $e$ maps $I$ surjectively onto $J$, or in other words $eI=J$.

(3) $I$ is the unique largest such ideal. I claim that if $K$ is an ideal such that $eK\subseteq J$, then $K\subseteq I$. This is because if $eK\subseteq J$, then for any element $k\in K$ we have that $ek\in J$, hence $k\in I$ by definition. This implies that any other ideal $K$ such that $eK=J$ is contained in $I$, hence $I$ is the unique largest ideal such that $eI=J$.

Suppose $J$ is prime but $I$ is not prime. Then there exist two elements $a,b$ such that $a,b\notin I$ but $ab\in I$. However, $ea\notin J$, and $eb\notin J$, and $(ea)(eb)=eab\in J$, contradicting the fact that $J$ is prime.

Answer 2

(1) does not hold in general. Take $R = \mathbb Z/ 6\mathbb Z$, idempotent $4$ and prime ideal $\langle 3 \rangle = \{ 0, 3\}$. Then $4\langle 3\rangle = \{0\}$ which is not prime in $\mathbb Z/ 6\mathbb Z$ because it's not integral domain.

But, in general, $eI$ is prime in $eR$, whenever $I$ is prime ideal in $R$ that doesn't contain idempotent $e$.

Let us elaborate.

Since $e$ is idempotent, $r\mapsto er$ is epimorphism of rings $R$ and $eR$. Thus, $eI$ is ideal as epimorphic image of an ideal $I$.

Now, for any epi $f\colon R\to S$, by first isomorphism theorem we have $R/f^{-1}(I) \cong S/I$ for any ideal $I$ in $S$, which gives us $R/f^{-1}(f(I)) \cong S/f(I)$ for any ideal $I$ in $R$. It is easy to check that $f^{-1}(f(I)) = I +\ker f$, thus $f(I)$ is prime in $S$ if and only if $I+\ker f$ is prime in $R$.

Let's set $f = r\mapsto er$ which is epimorphic map from $R$ to $eR$. If we prove that $\ker f\subseteq I$, then we have that $I = I +\ker f$ prime in $R$ implies that $eI$ is prime in $eR$. Assume that $I$ is prime and $e\notin I$. Now \begin{equation} (*)\quad\quad x\in \ker f\implies ex = 0\implies ex\in I \stackrel{e\notin I}{\implies} x\in I \end{equation} and thus, $\ker f\subseteq I$, which completes the proof.

We could prove this directly as well. Assume $(ex)(ey) \in eI$. Then, $exy = (ex)(ey) = ep$ for some $p\in I$. Now we have $$ e(xy - p) = 0 \implies xy - p \in\ker f\stackrel{(*)}{\implies} xy - p\in I$$ $$\implies xy\in I\implies x\in I\vee y\in I\implies ex\in eI\vee ey\in eI.$$

For (2), Matt Samuel already gave a good answer, details are easy to check, but if it's problematic for you, leave a comment, and I could help.