1. ## 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?

2. 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 $\displaystyle I_a$ be the ideal generated by $\displaystyle I$ and $\displaystyle a$. then $\displaystyle I_a=I+Ra.$ obviously $\displaystyle I \subseteq I_a$ and $\displaystyle a \in I_a - I.$ hence: $\displaystyle I \subsetneq I_a.$ similarly $\displaystyle I \subsetneq I_b,$ because $\displaystyle b \in I_b - I$ and

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

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

so $\displaystyle I_a=Rv, \ J=Rw,$ for some $\displaystyle v,w \in R.$ clearly $\displaystyle I_aJ=Rvw.$ finally $\displaystyle JI_a \subseteq I$ is a trivial result of the definition of $\displaystyle J. \ \ \ \Box.$

3. 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

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

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

now let $\displaystyle g_1=1_G$ and $\displaystyle r \in RG$ be in the augmentation ideal. then $\displaystyle r=\sum_{i=2}^nr_i(g_i - 1)=(-r_2-r_3 - \cdots - r_n)g_1 + r_2g_2 + \cdots r_ng_n.$ thus $\displaystyle r\sum_{i=1}^n g_i=0,$ by the Lemma. Q.E.D.

5. 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.