Let A be a finite commutative ring with unity.
1 Prove: Every nonzero element of A is either a divisor of zero or invertible
To prove this, we will use the fact that a nonzero element $a \in A$ is a divisor of zero if and only if there exists a nonzero element $b \in A$ such that $ab = 0$. We will also use the fact that a nonzero element $a \in A$ is invertible if and only if there exists an element $b \in A$ such that $ab = ba = 1$.
First, let's prove that every nonzero divisor of zero is not invertible. Suppose $a$ is a nonzero divisor of zero. Then there exists a nonzero element $b$ such that $ab = 0$. Assume for contradiction that $a$ is invertible. Then there exists an element $c$ such that $ac = ca = 1$. Multiplying both sides of $ab = 0$ by $c$ on the right, we get $abc = 0c$, which simplifies to $ab = 0$. But since $ac = 1$, we have $ab = 0$ implies $b = (ab)c = 0c = 0$, which contradicts the assumption that $b$ is nonzero. Therefore, we conclude that every nonzero divisor of zero is not invertible.
Next, let's prove that every nonzero non-divisor of zero is invertible. Suppose $a$ is a nonzero non-divisor of zero. This means that for all nonzero elements $b \in A$, we have $ab \neq 0$. We want to show that $a$ is invertible. Consider the set $S = \{a^k \mid k \geq 0\}$. Since $a$ is not a divisor of zero, we know that $S$ does not contain $0$. Also, since $a$ is finite and $S$ is a set of powers of $a$, $S$ must be finite.
Now let's consider the product $a^n = a^m$ for some positive integers $n > m$. We can multiply both sides of this equation by $a^m$ on the right to obtain $a^n a^{-m} = a^m a^{-m}$, which simplifies to $a^{n-m} = 1$. Since $S$ is finite, there must exist positive integers $n > m$ such that $a^{n-m} = 1$, which means that $a$ has a multiplicative inverse.
Therefore, we have shown that every nonzero element of a finite commutative ring with unity is either a divisor of zero or invertible.