A finite integral domain is a field
Dependencies:
Let $D$ be a finite integral domain. Then $D$ is a field.
Proof
Let $D^* = D - \{0\}$. Let $a \in D^*$. We must show that $a$ has a multiplicative inverse.
Let $\lambda_a: D^* \mapsto D^*$ where $\lambda_a(d) = ad$.
\begin{align} & \lambda_a(d_1) = \lambda_a(d_2) \\ &\Rightarrow ad_1 = ad_2 \\ &\Rightarrow a(d_1 - d_2) = 0 \tag{distributivity} \\ &\Rightarrow d_1 - d_2 = 0 \tag{$a \neq 0$ and $D$ is an integral domain} \\ &\Rightarrow d_1 = d_2 \end{align}
Therefore, $\lambda_a$ is one-to-one. Since the domain and co-domain of $\lambda_a$ have the same size, $\lambda_a$ is also onto.
Therefore, pre-image of 1 exists in $\lambda_a$. Let it be $d$. Therefore, $ad=1$. Therefore, inverse of $a$ exists (and it is $d$).
Dependency for: None
Info:
- Depth: 3
- Number of transitive dependencies: 4