For a commutative ring $R$ with unity, I am looking for an equivalent condition for an ideal $T$ to have the property that $T$ contains a unique maximal proper subideal, equivalently, the sum of proper subideals of $T$ is not equal to $T$.
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$\begingroup$ Related: mathoverflow.net/questions/291748/a-special-type-of-ideals $\endgroup$– Zach TeitlerFeb 12, 2018 at 7:46
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1$\begingroup$ Also related: math.stackexchange.com/questions/2644665/… . In particular, this answer implies that if R is a Noetherian local ring, an ideal has precisely one maximal subideal if and only if it is a nonzero principal ideal. $\endgroup$– Neil EpsteinFeb 12, 2018 at 16:56
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1$\begingroup$ I don't know if this is useful to you but a module for which "the sum of two proper submodules is not equal to the whole module" is called a hollow module, which is basically the dual notion to a uniform module. So you might be talking about a hollow ideals. $\endgroup$– rschwiebFeb 12, 2018 at 18:38
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1 Answer
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Claim: Let $T$ be an ideal in a commutative ring. If $T$ contains a unique maximal proper subideal, then $T$ is principal.
Proof: Indeed take any element outside the unique maximal proper subideal. The ideal generated by that cannot be a proper subideal and hence has to be $T$.
So, this suggests that this is a very special property. For instance, that maximal subideal would have to contain the square of any generator of $T$ and if those generators are not zero-divisors, then they are associate (i.e. they can only differ by a unit multiple). If any generator is a zero-divisor then every element in $T$ is a zero-divisor. I think you can continue this line of thought to derive more special properties when this holds.
There is also a partial converse:
Claim: Let $T$ be a non-zero principal ideal in a commutative ring. Then every proper subideal of $T$ is contained in an ideal that is maximal among proper subideals in $T$.
(By the usual proof of existence of maximal ideals)
Proof: Let $T\subseteq R$ be a principal ideal generated by $t\in T$, $T'\subsetneq T$ a proper subideal, and consider the set of ideals $I\subseteq T$ such that $T'\subseteq I$, $t\not\in I$ which is the same as the set of proper subideals of $T$ containing $T'$. If there is a chain of such ideals $\ldots\subseteq I_\lambda\subseteq \dots$ with $\lambda\in \Lambda$, then $t\not\in J:=\cup_{\lambda\in\Lambda}I_\lambda$ is an upper bound of the chain and hence the statement follows by Zorn's lemma.
However, of course, this does not mean that there is a unique one...
On the positive side, here is a situation where what you are hoping for holds:
Claim: If $R$ is a DVR, then the ideals in $R$ form a totally ordered set.
Proof: This follows essentially from the definition of a DVR.
So, I suppose one can conclude that this holds sometime, but for most ideals in most rings it does not.
answered Feb 12, 2018 at 7:25
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2$\begingroup$ Well to be entirely accurate one has to exclude the zero ideal from this... $\endgroup$ Feb 12, 2018 at 9:45
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3$\begingroup$ @PeterHeinig the second statement "Claim" of the OP "Let $T$ be a principal ideal... then $T$ contains [...] proper subideals" is false for $T=0$. To say that $\{0\}$ is maximal among proper subideals of $\{0\}$ is not vacuously true, it is false, because it says in particular that $\{0\}$ is a proper subset of $\{0\}$, which means the existence of an element in the empty set. $\endgroup$– YCorFeb 12, 2018 at 11:15
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2$\begingroup$ Actually the standard conclusion of Zorn's lemma gives a correct statement without excluding anything: any proper subideal (of a principal ideal) is contained in a maximal proper subideal $\endgroup$ Feb 12, 2018 at 13:45
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$\begingroup$ @მამუკაჯიბლაძე: How do you get the upper-bound without the exclusion? $\endgroup$ Feb 13, 2018 at 17:37
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1$\begingroup$ @მამუკაჯიბლაძე: I don't understand your argument. Are you saying that any subideal of a principal ideal is also principal? How about $R=k[x,y,z]$ and $(xy,xz)\subseteq (x)$? $\endgroup$ Feb 13, 2018 at 19:54