# Proof

(related to Theorem: Finite Integral Domains are Fields)

• By hypothesis, $$(R, + ,\cdot)$$ is an integral domain, which is finite.
• Let $R := \{0,a_1:=1,a_2, a_3, \ldots, a_n\}$ be the $n+1 < \infty$ elements of $R.$
• We will show that every nonzero element of $R$ is a unit, i.e. has a multiplicative inverse.
• Let $r \in R$ be given with $r\neq 0.$
• We build the set $R' = \{r \cdot 1, ra_2, ra_3, \ldots, ra_n\}.$
• Because $R$ is closed under multiplication "$\cdot$", we have $R'\subseteq R$.
• Since $R$ is an integral domain, $R'$ contains no zero divisors, i.e. for all elements of $r'\in R'$ we have $r'\neq 0.$
• Moreover, all elements $r'\in R'$ are distinct, because from $ra_i = ra_j$ for some $i \neq j,$ it follows $a_i = a_j$ from the cancelation law in integral domains.
• In particular $ra_i = 1$ for some $i \in \{1, \ldots, n\},$ so $r$ has an inverse.
• Altogether, it follows that every nonzero element $r\in R$ is a unit, thus $R$ is a field.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013
2. Lang, Serge: "Algebra - Graduate Texts in Mathematics", Springer, 2002, 3rd Edition