# quotient rings 3

• Apr 24th 2009, 12:25 PM
mpryal
quotient rings 3
If R is a finite integral domain, show that R is a field.

• Apr 24th 2009, 12:50 PM
Jhevon
Quote:

Originally Posted by mpryal
If R is a finite integral domain, show that R is a field.

do you know what an integral domain is? do you know what a field is? look at the definitions. you need only show that the set of nonzero elements of a finite integral domain forms a group with respect to multiplication.
• Apr 24th 2009, 04:07 PM
TheAbstractionist
Quote:

Originally Posted by mpryal
If R is a finite integral domain, show that R is a field.

Hi mpryal.

Pick a nonzero element $a$ in $R$ and consider the set $\{a^n:n\in\mathbb Z\}.$ This must be a finite set of nonzero elements; hence there exist positive integers $i,j$ with $i such that $a^i=a^j.$ Thus $0_R=a^j-a^i=a^i(a^{j-i}-1_R).$ Since $a^i\ne0_R$ and $R$ is an integral domain, $a^{j-i}-1_R=0_R.$ $\therefore\ a^{j-i}=1_R.$ $\therefore\ a^{j-i-1}$ is the multiplicative inverse of $a$ in $R$. (Note that $j-i-1\ge0.)$
• Apr 24th 2009, 09:39 PM
ThePerfectHacker
Quote:

Originally Posted by mpryal
If R is a finite integral domain, show that R is a field.

Let $R^{\times} = \{r_1,...,r_n\}$ where $r_1$ is identity.
For $r\in R^{\times}$ consider $rr_1,rr_2,...,rr_n$.
By pigeonhole it means $rr_j = r_1=1$ for some $j$.
Thus, $r$ has inverse.