# Ring Theory

• Nov 2nd 2008, 05:51 PM
robeuler
Ring Theory
Hi everyone
I am having trouble with this problem. It is number 6 in section 8.2 of Dummit and Foote if you have it.

R is an integral domain and suppose that every prime ideal is principal.
a) (I have done this) Assume that the set of ideals of R that are not principal is nonempty and prove that this set has a maximal element under inclusion.

b) (Where I am stuck) Let I be an ideal which is maximal with respect to being nonprincipal, and let a, b in R with

ab in I, but a not in I and b not in I (so I is not a prime ideal)

Let Ia be the ideal generated by (I,a) and Ib be the ideal generated by (I,b). Define J={r in R| rIa is a contained in I}

Prove that Ia=(v) and J=(w) (generated by v and w, respectively) are principal ideals in R with I properly contained in Ib contained in J and IaJ=(vw) contained in I.

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First with Ia, I cannot figure out how to express Ia as being generated by v. I also tried to suppose that it is not principal, but haven't gotten anywhere. Any ideas?
• Nov 2nd 2008, 06:53 PM
NonCommAlg
Quote:

Originally Posted by robeuler
Hi everyone
I am having trouble with this problem. It is number 6 in section 8.2 of Dummit and Foote if you have it.

R is an integral domain and suppose that every prime ideal is principal.
a) (I have done this) Assume that the set of ideals of R that are not principal is nonempty and prove that this set has a maximal element under inclusion.

b) (Where I am stuck) Let I be an ideal which is maximal with respect to being nonprincipal, and let a, b in R with

ab in I, but a not in I and b not in I (so I is not a prime ideal)

Let Ia be the ideal generated by (I,a) and Ib be the ideal generated by (I,b). Define J={r in R| rIa is a contained in I}

Prove that Ia=(v) and J=(w) (generated by v and w, respectively) are principal ideals in R with I properly contained in Ib contained in J and IaJ=(vw) contained in I.

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First with Ia, I cannot figure out how to express Ia as being generated by v. I also tried to suppose that it is not principal, but haven't gotten anywhere. Any ideas?

i guess your ring is commutative. let $I_a$ be the ideal generated by $I$ and $a$. then $I_a=I+Ra.$ obviously $I \subseteq I_a$ and $a \in I_a - I.$ hence: $I \subsetneq I_a.$ similarly $I \subsetneq I_b,$ because $b \in I_b - I$ and

$I \subseteq I+Rb=I_b.$ now let $r \in I,$ then $ra \in I$ and thus: $rI_a=rI +Rra \subseteq I.$ thus by definition: $r \in J.$ hence: $I \subseteq J.$ we also have: $bI_a=bI + Rba \subseteq I,$ because $ab \in I.$ thus $b \in J.$

hence $I_b=I+Rb \subseteq J.$ so: $I \subsetneq I_b \subseteq J.$ now since $I$ is maximal among non-principal ideals, every ideal which properly contains $I$ must be principal. in particular $I_a$ and $J$ are principal.

so $I_a=Rv, \ J=Rw,$ for some $v,w \in R.$ clearly $I_aJ=Rvw.$ finally $JI_a \subseteq I$ is a trivial result of the definition of $J. \ \ \ \Box.$
• Nov 2nd 2008, 07:16 PM
robeuler
It always seems so clear after you put it like that.

Here is the last one I am struggling with, if you have the patience.

Let G be a finite group {g1,g2,...,gn} and assume R is a commutative ring. If r is any element of the augmentation ideal of RG then r(g1+g2+...+gn)=0.

The hint is to use the fact that the augmentation ideal in group rings is generated by {g-1|g in G}.

I tried to look at the augmentation map directly, but i can't figure out a general case regarding any element of the augmentation ideal. It doesn't help that my notion of the group ring is weak... but that is slowly improving.

thanks again
• Nov 2nd 2008, 08:02 PM
NonCommAlg
Quote:

Originally Posted by robeuler
It always seems so clear after you put it like that.

Here is the last one I am struggling with, if you have the patience.

Let G be a finite group {g1,g2,...,gn} and assume R is a commutative ring. If r is any element of the augmentation ideal of RG then r(g1+g2+...+gn)=0.

The hint is to use the fact that the augmentation ideal in group rings is generated by {g-1|g in G}.

I tried to look at the augmentation map directly, but i can't figure out a general case regarding any element of the augmentation ideal. It doesn't help that my notion of the group ring is weak... but that is slowly improving.

thanks again

this is a nice, although quite easy, problem! first we need a simple lemma:

Lemma: let $G=\{g_1, \cdots , g_n \}$ be a group, R a commutative ring, and $r=\sum_{i=1}^nr_ig_i \in RG.$ suppose that $\sum_{i=1}^n r_i = 0.$ then $r\sum_{i=1}^n g_i=0.$

Proof: clearly for any $1 \leq k \leq n$ we have: $G=\{g_1, \cdots , g_n \}=\{g_kg_1, \cdots , g_kg_n \}.$ thus: $r\sum_{i=1}^n g_i=(\sum_{i=1}^n r_ig_i)(\sum_{i=1}^n g_i)=\sum_{k=1}^n r_k\sum_{i=1}^n g_kg_i=\sum_{i=1}^n(\sum_{k=1}^nr_k)g_i = 0. \ \ \Box$

now let $g_1=1_G$ and $r \in RG$ be in the augmentation ideal. then $r=\sum_{i=2}^nr_i(g_i - 1)=(-r_2-r_3 - \cdots - r_n)g_1 + r_2g_2 + \cdots r_ng_n.$ thus $r\sum_{i=1}^n g_i=0,$ by the Lemma. Q.E.D.
• Nov 2nd 2008, 08:31 PM
robeuler
Thanks again!

I should have mentioned, I actually proved the lemma in part a of the exercise.

And it is a nice problem I agree. Dummit and Foote has a very nice collection of problems and examples.