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