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

Please show steps. Thanks!

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- Apr 24th 2009, 11:25 AMmpryalquotient rings 3
If R is a finite integral domain, show that R is a field.

Please show steps. Thanks! - Apr 24th 2009, 11:50 AMJhevon
- Apr 24th 2009, 03:07 PMTheAbstractionist
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 $\displaystyle j-i-1\ge0.)$ - Apr 24th 2009, 08:39 PMThePerfectHacker
Let $\displaystyle R^{\times} = \{r_1,...,r_n\}$ where $\displaystyle r_1$ is identity.

For $\displaystyle r\in R^{\times}$ consider $\displaystyle rr_1,rr_2,...,rr_n$.

These must be distinct.

By pigeonhole it means $\displaystyle rr_j = r_1=1$ for some $\displaystyle j$.

Thus, $\displaystyle r$ has inverse.