I need some help:
Prove that if an ideal I of R contains a unit then I=R
Prove that a commutative ring is a field iff it it contains exactly 2 ideals
Prove that if the ring R is commutative and local then R*=R\I
some hints/tips would be great.
Thanks
Let F be a field.
If , then it is ok. Suppose be an ideal of F. Then A contains a unit. From the previous, A = F.
Lemma: Let R be a commutative ring with unity. and M a maximal ideal of R. Then R/M is a field iff M is maximal.(You can show this.. Ü)
Suppose, F contains exactly 2 ideals. Since is the only proper ideal of F, then is a maximal ideal, and so is a field which is isomorphic to F. Thus, F is a field. QED.
Let R be a commutative ring (with unity) that contains only two ideals, {0} and R. Let I be an ideal of R. If R is a ring such that (kalagota forgot to mention that) then choose . Now define . Then clearly this is an ideal of R and is non-trivial so it means which means there exists a so that . This shows all the non-zero elements of R has inverses, this makes R a field. Q.E.D.
can you help me with this part:
Prove that if the ring R is commutative and local then R*(units of R)=R\I, where I is the largest proper ideal of R?
it seems to me that it follows from the last two parts of the question but i can't phrase it correctly.
Thanks