# Thread: Algrbra Help!!! rings and ideals :(

1. ## Algrbra Help!!! rings and ideals :(

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

I need some help:

Prove that if an ideal I of R contains a unit then I=R
Let $u \in I$, where u is a unit.
Then there is $u^{-1} \in R$ such that $u^{-1}u =1$. Since, I is an ideal, $1 \in I$
Let $r \in R$. Then $r \cot 1 \in I$. Thus $R \subseteq I$. Therefore, I = R..

I need some help:

Prove that a commutative ring is a field iff it it contains exactly 2 ideals
Let F be a field.
If $A = \left\{ 0 \right\}$, then it is ok. Suppose $A \not = \left\{ 0\right\}$ 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 $\left\{ 0 \right\}$ is the only proper ideal of F, then $\left\{ 0 \right\}$ is a maximal ideal, and so $F/\left\{ 0 \right\}$ is a field which is isomorphic to F. Thus, F is a field. QED.

4. 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 $1\not = 0$ (kalagota forgot to mention that) then choose $a\not = 0$. Now define $\left< a\right> = \{ ra|r\in R\}$. Then clearly this is an ideal of R and is non-trivial so it means $\left< a \right> = R$ which means there exists a $b$ so that $ab=1$. This shows all the non-zero elements of R has inverses, this makes R a field. Q.E.D.

5. ## can you help me with the third part of the question?

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

6. Originally Posted by ThePerfectHacker
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 $1\not = 0$ (kalagota forgot to mention that) then choose $a\not = 0$. Now define $\left< a\right> = \{ ra|r\in R\}$. Then clearly this is an ideal of R and is non-trivial so it means $\left< a \right> = R$ which means there exists a $b$ so that $ab=1$. This shows all the non-zero elements of R has inverses, this makes R a field. Q.E.D.
maybe, or maybe not (did i forget). It's because i used to prove the exercise using the lemma i stated.. anyways, thanks TPH!!