Chủ đề này có 5 trả lời

[nemo]

Can someone give me an example of a ring with no maximal ideals.

We know that by a Zorn’s lemma argument, a ring with identity has a maximal ideal. The most common example in textbooks of a ring with no maximal ideals is to tart with the group Z(p∞), which is the subgroup of Q/Z of elements of order a power of the prime p. I want find another simple examples without using discrete valuation ring (because by this way, i have examples of P.Morandi).

[Ham_Toan]

Can someone give me an example of a ring with no maximal ideals.

I think a ring always had at least a maximal ideal.

[nemo]

I think a ring always had at least a maximal ideal

You can proceed with the group Z(p∞), which is the subgroup of Q/Z of elements of order a power of the prime p. It is known that the subgroups of Z(p∞) form an infinite increasing chain, and so there is no maximal subgroup. By defining multiplication in Z(p∞) by x · y = 0, this group becomes a ring with no maximal ideals.

[Ham_Toan]

I think a ring always had at least a maximal ideal

You can proceed with the group Z(p∞), which is the subgroup of Q/Z of elements of order a power of the prime p. It is known that the subgroups of Z(p∞) form an infinite increasing chain, and so there is no maximal subgroup. By defining multiplication in Z(p∞) by x · y = 0, this group becomes a ring with no maximal ideals.

I don't understand your explanation, can you explain it more clearly ?
Thank you so much ?

[canh_dieu]

Nemo’s example is based on the following observation: If http://dientuvietnam...n/mimetex.cgi?G with a ring structure by defining http://dientuvietnam...metex.cgi?a*b=0 for all http://dientuvietnam...n/mimetex.cgi?G are then precisely subgroups of http://dientuvietnam.../mimetex.cgi?G. Thus if we can find an abelian group G whose subgroups form an infinite increasing chain then http://dientuvietnam...n/mimetex.cgi?G considered as a ring has no maximal ideal. The group http://dientuvietnam.net/cgi-bin/mimetex.cgi?Z(p^{\infty}) is such a candidate.

First let us try to understand what http://dientuvietnam.net/cgi-bin/mimetex.cgi?Z(p^{\infty}) is the subgroup of the quotient group http://dientuvietnam...mimetex.cgi?Q/Z consisting of elements of order a power of http://dientuvietnam.../mimetex.cgi?p. Let http://dientuvietnam...x.cgi?[a/b] be an element of http://dientuvietnam.net/cgi-bin/mimetex.cgi?Q/Z). Saying that this element has order a power of http://dientuvietnam.net/cgi-bin/mimetex.cgi?p is equivalent to saying that there is http://dientuvietnam.net/cgi-bin/mimetex.cgi?p^n[a/b]=0 in http://dientuvietnam.net/cgi-bin/mimetex.cgi?Q/Z, or that http://dientuvietnam.net/cgi-bin/mimetex.cgi?b divides http://dientuvietnam.net/cgi-bin/mimetex.cgi?p^n. Hence we can easily see that http://dientuvietnam.net/cgi-bin/mimetex.cgi?Z(p^{\infty}) consists of elements of the form http://dientuvietnam.net/cgi-bin/mimetex.cgi?W be a proper subgroup of http://dientuvietnam.net/cgi-bin/mimetex.cgi?a with http://dientuvietnam.net/cgi-bin/mimetex.cgi?l such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?p that appear in the denominators of reduced fractions representing elements of http://dientuvietnam.net/cgi-bin/mimetex.cgi?W is bounded. This is just an easy consequence of the above fact.

Therefore for each nonzero proper subgroup http://dientuvietnam.net/cgi-bin/mimetex.cgi?W of http://dientuvietnam.net/cgi-bin/mimetex.cgi?n such that It is now clear that the proper subgroups of form an infinite increasing chain.

Bài viết đã được chỉnh sửa nội dung bởi canh_dieu: 13-11-2005 - 05:27

[madness]

The explanation of canh_dieu is pretty clear. However, many authors consider a ring as a "ring with a unit", and they just call a ring without unit as a "nonunital ring". So, to reformulate the original question, we need to find a nonunital ring without any maximal ideal.

Should this also be an example? Let A be a nonunital ring, whose underlying set is the direct sum of the abelian groups Z/p, where p runs through the set of all prime numbers, and whose multiplication is the one trivially endowed as mentioned above. Then it has no maximal ideal, because if I is an ideal of A and x is in I, then for all the nonzero coordinates of x, we easily prove that those corresponding direct summands of A also lie completely in I. For example, if x is (0 mod 2, 1 mod 3, 3 mod 5, 0 mod 7, 0 mod 11, ...) then the whole direct summand Z/3 and Z/5 lie in I. Hence for any proper ideal of A, we can enlarge it without making it nonproper.