Showing an ideal with maximality condition is prime.

Question

Let $R$ be a commutative domain and suppose that $I \subseteq R$ is an ideal of $R$ maximal with respect to the property that $I^{-1} \not\subseteq R$. Show that $I$ is a prime ideal.

This is deemed a harder homework question by my instructor. Some of my thoughts are as below.

Notationally, if $J$ is an ideal of $R$ and $K$ is the field of fractions for $R$, then $$J^{-1} = \{k \in K \mid kJ \subset R \}$$

Begining of possible proof:

Take $r,s$ in $R$. Suppose that that their product $rs$ is in $I$, but that $r$ is not in $I$. Then we have that the following ideal of $R$ properly contains $I$: $$\langle r \rangle + I$$ where $ \langle r \rangle$ is the ideal generated by $r$ in $R$. We know $$I \subsetneq (\langle r \rangle + I) \subseteq R.$$

Note that clearly $\langle r \rangle^{-1} = \left\{\dfrac{z}{r}\mid z \in R\right\}$.

If $r = 1_R$, then $\langle r \rangle^{-1} = \langle r \rangle = R$. Otherwise, $\langle r \rangle^{-1} \not\subseteq R$. It should be relatively evident that $(\langle r \rangle + I)^{-1} = \langle r \rangle^{-1} \cap I^{-1}$. This can be thought of by considering that the only things that will invert both $\langle r \rangle$ and $I$ will lie in the intersection of the two ideal inverses.

I perhaps want to use this to somehow show that $(\langle r \rangle + I)^{-1} \not\subseteq R$, which by maximality would imply that $(\langle r \rangle + I) = R$. This would allow me to write $(rx) + a = 1$ for $rx \in \langle r \rangle, a \in I$.

Then $s(rx) + sa =(sr)x + sa= s$. We know that $sr \in I$, so $(sr)x \in I$. Also, by absorption, $sa$ is in $I$. So $s$ must be in $I$ and we are done.

The one thing I cannot see how to show is that $(\langle r \rangle + I) = R$. Any thoughts?

Answer 1

Picking $I \subseteq R$ to be maximal with respect to this condition, you would then have that for $0\neq r\in R-I$, $I\subsetneq (r)+I$. Thus we must have $((r)+I)^{-1}\subseteq R$. So what you're trying to show is a contradiction to the very condition you are starting with.

Instead, consider that $((r)+I)^{-1}=R$. This means that the set $\{\dfrac{z}{r}\vert z\in R-(r)\}\bigcap I^{-1}=\emptyset$. But we're assuming that $rs\in I$ and $\exists k\in I^{-1}-R$; now all such $k$ have the property that $krs\in R$, so in particular, any "denominator" in $k$ must be cleared out by $rs$. But it can't be cleared out by r.

Edit

I guess what I was trying to say was that the proof runs cleaner without shooting for that contradiction. Let $I\subsetneq R$ be maximal with respect to this condition, and suppose we have $r,s\in R$ such that $rs\in I$ and $r\notin I$. Then $((r)+I)$ is an ideal of $R$ and $I\subsetneq ((r)+I)$, whence by maximality we have $((r)+I)^{-1}=R$. Now, as noted above $((r)+I)^{-1}=(r)^{-1}\bigcap I^{-1}=\{\dfrac{z}{r}\vert z\in R\}\bigcap \{k\in K\vert kI\subseteq R\}$.

Picking $k\in I^{-1}-R$ by the above we see that $kr\notin R$ and $krs\in R$. Thus $kr=\frac{z}{s}$ for some $z\in R-(s)$. Then $((s)+ I)^{-1}\not\subseteq R$. Now as $I$ is maximal with respect to $I^{-1}\not\subseteq R$, and $I\subseteq ((s)+I)$, we must have equality, whence $s\in I$, so I is prime.

Answer 2

This is a possible answer that I have come up with. It can possibly be trimmed and edited for clarity.

Let $R$ be a commutative domain and suppose that $A \subseteq R$ is an ideal maximal with respect to the property that $A^{-1} \not\subseteq R$. We show that $A$ must be prime.

Take $r,s$ in $R$. Suppose that that their product $rs\in A$, but that neither $r$ nor $s$ are in $A$. Then we have that the following ideals of $R$ properly contains $A$: $$\langle r \rangle + A$$ where $ \langle r \rangle$ is the ideal generated by $r$ in $R$ and $$\langle s \rangle + A$$ where $ \langle s \rangle$ is the ideal generated by $s$ in $R$. We know $$A \subsetneq (\langle r \rangle + A) \subseteq R.$$ $$A \subsetneq (\langle s \rangle + A) \subseteq R.$$

It should be relatively evident that $(\langle r \rangle + A)^{-1} = \langle r \rangle^{-1} \cap A^{-1}$, and similarly for $(\langle s \rangle + A)^{-1}$. This can be thought of by considering that the only things that will invert both $\langle r \rangle$ and $A$ will lie in the intersection of the two ideal inverses.

Since both $\langle r \rangle + A$ and $\langle s \rangle +A$ strictly contain $A$, if either $(\langle r \rangle + A)^{-1}$ or $(\langle s \rangle +A)^{-1}$ is not contained in $R$ we will have a contradiction of maximality of $A$ unless the corresponding ideal sum ($\langle r \rangle + A$ or $\langle s \rangle +A$) equals $R$. Suppose that neither $\langle r \rangle + A$ nor $\langle s \rangle +A$ equals $R$. Hence $$(\langle r \rangle + A)^{-1}=(\langle s \rangle+A)^{-1}=R$$ since both $(\langle r \rangle + A)$ and $(\langle s \rangle +A)$ are $R$-modules contained in $R$.

Take $k \in A^{-1}$ with $k \not\in R$ (Such a $k$ must exist since $A^{-1}$ is not contained in $R$). Since $rs \in A$, we have $krs \in R$. Furthermore, since $A^{-1}$ is an $R$-module, $kr \in A^{-1}$.

Suppose that $kr \notin R$. We know $(kr)s = krs \in R$, so $kr \in \langle s \rangle^{-1}$. But then this would imply that $kr \in (\langle s \rangle +A)^{-1}$ since $kr$ is in both $\langle s \rangle^{-1}$ and $A^{-1}$. This would contradict $(\langle s \rangle +A)^{-1}$ being equal to $R$. So in fact we must have that $kr \in R$. Since $k$ was arbitrary in its role, we will have that for \emph{any} $k \in A^{-1}$, $kr \in R$.

Aside: We took arbitrary $k$ in $A^{-1}$ not in $R$ and showed that $kr \in R$ for any such $k$. Clearly we have as well that $kr \in R$ when $k$ is in $R$ since $\langle r \rangle$ is an $R$-module.

But then this tells us that $A^{-1}\langle r \rangle \subseteq R$. So then in fact $(\langle r \rangle + A)^{-1} = A^{-1} \not\subseteq R$. By maximality of $A$, the only case we can have for $(\langle r \rangle + A)$ is equality to $R$. Thus $(\langle r \rangle + A) = R$ is a must.

Since $(\langle r \rangle + A) = R$, then there exists $(rx) \in \langle r \rangle$ and $y \in A$ such that $rx + y = 1_R$. Then we have $$s(rx + y) = s(rx) + sy = (sr)x + sy = s.$$ Since $sr \in A,$ then $(sr)x \in A$. Since $y \in A$, then $sy \in A$. Hence $s$ itself is in $A$. Thus $A$ is prime.