# Does minimal ideal always imply principal ideal?

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:

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

• (2) is not correct. That $I$ is principal does not mean $I = (y)$ for every $y \in I$, it means $I = (y)$ from some $y \in I$.
– Jim
Oct 6, 2019 at 13:36
• (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.
– Jim
Oct 6, 2019 at 13:37
• 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. Oct 6, 2019 at 13:37
• Oh, actually you're right! I did read what you were proving incorrectly, my bad.
– Jim
Oct 6, 2019 at 13:39
• In that case both (1) and (2) are correct.
– Jim
Oct 6, 2019 at 13:39

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.

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 $$R$$.

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.)

• Yes, it does. (You've proved it, right?)
– Orat
Oct 6, 2019 at 13:42
• 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 .. . Oct 6, 2019 at 13:47
• 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.
– Orat
Oct 6, 2019 at 13:53
• 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... Oct 6, 2019 at 13:56
• 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 $xR$.
– Orat
Oct 6, 2019 at 14:02