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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:

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

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    $\begingroup$ (2) is not correct. That $I$ is principal does not mean $I = (y)$ for every $y \in I$, it means $I = (y)$ from some $y \in I$. $\endgroup$
    – Jim
    Oct 6, 2019 at 13:36
  • $\begingroup$ (1) is correct though, all minimal ideals are principal and for minimal ideals it actually is true that $I = (y)$ for every $y \in I$. That condition is actually equivalent to the principal ideal being minimal. $\endgroup$
    – Jim
    Oct 6, 2019 at 13:37
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    $\begingroup$ Then you have read incorrectly. I'm not saying that I is principal. I'm saying that I is an ideal which has the property that it is generated by any of its elements. $\endgroup$
    – AfterMath
    Oct 6, 2019 at 13:37
  • $\begingroup$ Oh, actually you're right! I did read what you were proving incorrectly, my bad. $\endgroup$
    – Jim
    Oct 6, 2019 at 13:39
  • $\begingroup$ In that case both (1) and (2) are correct. $\endgroup$
    – Jim
    Oct 6, 2019 at 13:39

3 Answers 3

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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.

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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 $R$.

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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.)

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  • $\begingroup$ Yes, it does. (You've proved it, right?) $\endgroup$
    – Orat
    Oct 6, 2019 at 13:42
  • $\begingroup$ Well, I'm not sure, because I can't find the statement mentioned on the web. On the wikipedia page of minimal ideal, en.wikipedia.org/wiki/Minimal_ideal , they mention briefly that for a ring R with unity, this in neccesarily true for right ideals .. . $\endgroup$
    – AfterMath
    Oct 6, 2019 at 13:47
  • $\begingroup$ Well, what you've done is basically the same as the argument shown on that wikipedia page. Depending on what type of ideals you're considering (right/left/two-sided), considering an ideal of the form ($xR$/$Rx$/$RxR$) is the key, anyway. BTW, many people may reserve the symbol $(x)$ to denote $RxR$ only. $\endgroup$
    – Orat
    Oct 6, 2019 at 13:53
  • $\begingroup$ Well, the fact that I=xR / I=Rx / I=RxR for principal ideals is only the case if R has unity, which isn't assumed... $\endgroup$
    – AfterMath
    Oct 6, 2019 at 13:56
  • $\begingroup$ You're right; as I usually deal with unital rings, I didn't pay much attention to that. Anyway the former part of your proof is correct as it only uses the minimality, and does not use something like $xR$. $\endgroup$
    – Orat
    Oct 6, 2019 at 14:02

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