quotient rings 3

**mpryal** · April 24th 2009, 11:25 AM

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

Please show steps. Thanks!

**Jhevon** · April 24th 2009, 11:50 AM

Originally Posted by mpryal

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

Please show steps. Thanks!

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.

**TheAbstractionist** · April 24th 2009, 03:07 PM

Originally Posted by mpryal

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

Please show steps. Thanks!

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 $i<j$ 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 $j-i-1\ge0.)$

**ThePerfectHacker** · April 24th 2009, 08:39 PM

Originally Posted by mpryal

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

Please show steps. Thanks!

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

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