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Is there any equivalence condition for a commutative ring to have exactly two or three maximal ideals?

user26857
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arenna
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    Not sure what you are after here exactly, but such rings are easy to construct. For example, consider the ring obtained from $\mathbb Z$ by inverting all integers relatively prime to $6$. This ring has two maximal ideals, those generated by $2$ and by $3$. – John Brevik Jun 25 '15 at 12:07
  • @JohnBrevik That one is particularly interesting since it's also a domain, and therefore not semiperfect. The four idempotents of $R/J(R)$ can't lift to the two idempotents in $R$. – rschwieb Jun 25 '15 at 14:01
  • Right. A similar thing happens when you take the integral closure of the local ring of a node on a curve. – John Brevik Jun 26 '15 at 16:45
  • @JohnBrevik How can you prove that there is no maximal ideals other than $(2)$ and $(3)$? – user Mar 16 '16 at 14:24
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    @user The ideals of that ring are precisely the extended ones from $\mathbb Z$, and all other primes meet the multiplicatively closed set, so they become the unit ideal in the ring of fractions. – John Brevik Mar 16 '16 at 18:34

2 Answers2

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I don't know if this could be usefull: I give to you some tricks to compute the number of maximal ideals of some kind of rings.

Consider the Jacobson radical $J(A)$ of a semilocal ring $A$ with maximal ideals $M_1, \dots , M_n$, then
$$ A/J(A) \simeq A/M_1 \oplus \dots \oplus A/{M_n} $$

Now, compute the number of factors of $$ A/M_1 \oplus \dots \oplus A/M_n$$ is easy by looking at the primitive idempotents (I think there are serveral algorithms that do this, for example looking at the Kernel of the map that sends $x$ to $x^2$). If $A$ is a finitely generated $k$ algebra, with $k$ algebrically closed, you have just to look at the $rank$ of $A/J(A)$ as $k$-module.

As an easy consequence of this fact, we can easy look at number of factors of a polynomial $p$ of $k[x]$:

If $p=\prod_1^n p_i$ (we can easy reduce to the case of a squarefree polynomial considering $p/GCD(p,p')$) we can consider

$$k[x]/(p) \simeq k[x]/(p_1) \oplus \dots \oplus k[x]/(p_n) $$

and count the number of idempotents in the $k$ algebra above!

For some algorithms to compute idempotents you can look at P.Gianni,V. Miller,B. Trager - Decomposition of Algebras

Other ways:

If $A$ is a finetely generated $k$-alebra (say $k[x_1, \dots , x_n]/I$), then the number of its maximal ideals is the number of closed points of $V(I)$.

To find the number of this closed points, compute the reduced Grobner base $RG$ of $I$ and count the number of lattice points in the sub-escalier of $RG$.

You can find a better description of this metod in my answer Here

Community
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Sabino Di Trani
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  • The maximal ideals of $A$ correspond exactly to those of $A/J(A)$, and since the latter is just a product of fields ($n$ of them, say) then it is elementary that there are $n$ maximal ideals. – rschwieb Jun 25 '15 at 13:00
  • You're reasoning is trivial "a posteriori", if you "a priori" don't know anyting on the number of Maximal ideals but know someting about the jacobson radical (as in the the case of $k[x]/(p)$), well just look at $A/J(A)$ and compute its primitive idempotents. – Sabino Di Trani Jun 25 '15 at 13:11
  • Observe that, if you don't know the number of maximal ideals, you can't compute the number of fields in the product.... – Sabino Di Trani Jun 25 '15 at 13:14
  • I didn't realize you were looking for a way to determine $n$, sorry. I took the decomposition for granted. Regards – rschwieb Jun 25 '15 at 13:24
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The most straightforward characterization that comes to mind is this:

A commutative ring $R$ has only two (resp three, or $n$) maximal ideals iff $R/J(R)$ is isomorphic to the direct product of two (resp three, or $n$) fields. ($J(R)$ denotes the Jacobson radical of $R$.)

I guess you are probably looking for something along the lines of the nice characterization of local rings in terms of elements. I have not heard of a nice counterpart for more maximal ideals, not even the special cases of $2$ or $3$.

There is only one nice theorem I have ever heard regarding rings having exactly two maximal ideals:

The endomorphism ring of a nonzero uniserial right module (one in which the submodules form a chain) has exactly one or exactly two maximal right ideals.

If I recall correctly, one of the ideals is comprised totally of non-surjective endomorphisms, and the other is non-injective endomorphisms (see page 22 for example. It can also be found in Lam's First Course in Noncommutative Rings) It is problematic, of course, to know when these endomorphism rings would be commutative.

rschwieb
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