Does minimal ideal always imply principal ideal?

Question

First, let me specify two definitions i will use.

$[1.]$ A (right/ left/ both) ideal $I$ of a ring $R$ (unity not assumed) is minimal if $(1.) \; I\neq (0)$ and $(2.)$ If $J$ is any nonzero (right/ left/ both) ideal of $R$ containied in $I$, then $J=I$

$[2.]$ If $x \in R$, then $(x)$ is the intersection of all (left/ right/ both) ideals of $R$ containing $x$.

Consider the following propostition and its proof:

$\textbf{A [right / left / both] ideal $I$ of a ring $R$ is minimal iff}$

$ \textbf{ $I$ is generated by any of its nonzero elements $x \in I$ }$

Proof:

$1.(\Rightarrow)$ Suppose $I$ is minimal and and $x \in I$ is nonzero. Consider the ideal $J:=(x)$ generated by $x$. By construction, $J \neq (0)$ since $x \in J$. Now, $J \subseteq I$, since by definition $J$ is the smallest ideal containing $x$. But then , by minimality of $I$ we must have $I=J$, so $I$ is generated by $x$.

$2.(\Leftarrow)$ Suppose $I=(x)$ for any $x \in I$, and that $J$ is any nonzero ideal of $R$ with $J \subseteq I$. Let $y$ be any nonzero element of $J$. Then $y \in I$, and by hypothesis we have $I=(y)$. But then we must have $J=I$, because $(y)$ is the smallest ideal of $R$ containing $y$.

My question is:

$\textbf{Does this also prove that any minimal ideal is principal? }$

Answer 1

Yes essentially, although I think it is safest to word it as “a left (resp, right/twosided) ideal is minimal if and only if it is generated by any of its nonzero elements as a left (resp, right/twosided) ideal.”

The fact that minimals are generated by a single element follows a fortiori from the $\implies$ direction.

Answer 2

I agree with the former part of your proof, but I do not with the latter part. (I missed the word ANY.)

To be specific, consider $R = \mathbb Z$ and an ideal $I = (n)$ with $n \neq 0$. Obviously, it contains a non-zero proper ideal $$(n) \supset (2n) \supset (0).$$ Hence, a principal ideal generated by a non-zero element needs not to be minimal. (Indeed, this argument shows that $\mathbb Z$ has no minimal ideals.)

Answer 3

Let $I$ be a minimal right ideal of a ring $R$. By definition, $I$ being a principal right ideal means that $\exists x \in R \, (I=xR)$.

In fact, $\forall x \in I \setminus \{0\} \, (I=xR)$. Indeed, for any nonzero element $x$ of $I$, $xR \subseteq I$ (because $I$ is a right ideal of $R$) and $0 \neq x \in xR$, so $xR=I$ because $I$ is assumed to be a minimal right ideal.

Similarly, if $I$ is a minimal left (resp. two-sided) ideal of $R$, then $\forall x \in I \setminus \{0\} \, (I=Rx)$ (resp. $\forall x \in I \setminus \{0\} \, (I=RxR)$).

More generally, any simple module is cyclic. Two-sided ideals of $R$ are the same as the right $R^{op} \otimes_{\mathbb{Z}} R$-submodules of $R$.