How to determine if this is a field?

Question

A finite subring $R$ of a field $V$ contains $1$ (so $1$ is an element of $R$).

The question is: True or False: The ring $R$ must be a field.

I thought that if $R$ was a field it had to be a finite field in this case (because $R$ is a finite ring). And to be able to be a finite field, it should have $$ |R| = p^n $$ With $p$ a prime and $n$ a natural number, but this doesn't have to be the case.

I could be wrong or what I'm saying could be insufficient to prove this right or wrong.

Please help ^^

Answer 1

Hint: It is a finite integral domain. Or else if you are not acquainted with the relevant theorem, the only issue is whether every non-zero element $a$ of $R$ has a (multiplicative) inverse.

To show that it does, consider the powers of $a$. By finiteness, there must be integers $m$ and $n$, with $0\le m \lt n$ such that $a^m=a^n$. But then $a^{n-m}=1$. From this you should be able to show that some power of $a$ is the inverse of $a$.

Remark: You are right that if $R$ is a field (which it is) then $R$ will have $p^n$ elements for some prime $p$ and some positive integer $n$. However, that does not lead to a contradiction.

Answer 2

Since $R$ is a finite subring, it is sufficient to check whether it is an integral domain, as every finite integral domain is a field.

Of course, a (sub)ring is an integral domain if it has no zero divisors.

Assume by contradiction that $R$ has zero divisors. Then, there are $a,b \in R$ such that $ab = 0$.

$V$ is a field, and so $V$ is an integral domain and has no zero divisors. Since $a,b \in R \subset V$, then $V$ has zero divisors, contradicting the statement that $V$ is a field.