1. ## quotient rings 3

If R is a finite integral domain, show that R is a field.

2. 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.

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

Hi mpryal.

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

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

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