# Thread: Condition for Commutative Ring to be a Field.

1. ## Condition for Commutative Ring to be a Field.

Hi,

Suppose $R \neq \{0\}$ is a commutative ring with no zero divisors.
Suppose that for every $a \in R$, $a \neq 0$, we have $aR = R$.

Prove R is a field.

I am not sure what to do... just need to show that each $a \in R$ has an inverse (ie, each element is a unit)? Not sure how to go about it.

Thanks!

2. I suppose we know that $\ = R\ , \forall a \in R$...

3. I think I made some progress? not sure if it is valid...

Since $\ = R$, then we can write each element $r \in R$ as $r = ac$ for some $c \in R$.

But since $a$ itself is in $R$, then $ac = a$ for some $c \in r$.

Since $R$ is commutative, $ac = ca = a$
Does this mean that we have a unity? How do we prove that $c$ works for every element of $R$?

Once we can show we have a unity, I think the rest is done, because:
if $r \in R$, then $ = R$. So each element of $R$ can be written as $rd$ for some $d \in R$. Since $1_R \in R$, $1_R = rd$ for some $d \in R$. Since $R$ is commutative, $1_R = rd = dr$, so by the definition of unique inverses, each element has an inverse, so each element is a unit, so $R$ is a field.

Can anyone look over this argument and verify it/spot a problem? I'd be very grateful, thanks!

4. Originally Posted by matt.qmar
I think I made some progress? not sure if it is valid...

Since $\ = R$, then we can write each element $r \in R$ as $r = ac$ for some $c \in R$.

But since $a$ itself is in $R$, then $ac = a$ for some $c \in r$.

Since $R$ is commutative, $ac = ca = a$
Does this mean that we have a unity? How do we prove that $c$ works for every element of $R$?

Once we can show we have a unity, I think the rest is done, because:
if $r \in R$, then $ = R$. So each element of $R$ can be written as $rd$ for some $d \in R$. Since $1_R \in R$, $1_R = rd$ for some $d \in R$. Since $R$ is commutative, $1_R = rd = dr$, so by the definition of unique inverses, each element has an inverse, so each element is a unit, so $R$ is a field.

Can anyone look over this argument and verify it/spot a problem? I'd be very grateful, thanks!

1) $aR=R\Longrightarrow \exists\,c_a\in R\,\,s.t.\,\,ac_a=a$

2) $aR=R\Longrightarrow \forall\,x\in R\,\,\exists\, r_x\in R\,\,s.t.\,\,ar_x=x$

3) $\forall\,x=ar_x\in R\,,\,\,xc_a=(ar_x)c_a=r_x(ac_a)=r_xa=x\Longrighta rrow c_a$ is the

whole ring's multiplicative unit.

Tonio