Let $A$ a finite ring. Prove that $A$ is a field, or $A$ has zero divisors. Let $A$ a finite ring. Prove that $A$ is a field, or $A$ has zero divisors.

I begin to assume that $A$ has no zero divisors but I don't know continue...
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How would be this proof? thanks! :)
 A: Since $A$ is finite (say, $|A| = n$), we can list its $n$ distinct elements as:
$$
A = \{0, a_2, a_3, \ldots, a_n\}
$$
Now choose any nonzero $x \in A$. We claim that $x$ is a unit, so that $A$ is a field. To this end, consider the set:
$$
B = \{xa_2, xa_3, \ldots, xa_n\}
$$
Since $A$ has no zero divisors, we know that these $n - 1$ elements are all nonzero (so that $B \subseteq A \setminus \{0\}$). In fact, we claim that they are all distinct (so that $|B| = n - 1 =  |A \setminus \{0\}|$).


*

*Otherwise, suppose that $xa_i = xa_j$ for some $2 \leq i < j \leq n$. Then $x(a_i - a_j) = 0$. But then since $x \neq 0$ and $A$ has no zero divisors, we know that $a_i - a_j = 0$ so that $a_i = a_j$, a contradiction.


Thus, it follows that $B = A \setminus \{0\}$ so that $1 \in B$. But then it follows that $xa_k = 1$ for some $k \in \{2, 3, \ldots, n\}$. So $x$ is a unit, as desired. $~~\blacksquare$
A: Suppose $A$ has no zero divisors, and for any $0 \ne a \in A$, consider the map $\theta_a: A \to A$ given by $\theta_a(r) = ar$ for $r \in A$.  If $\theta_a(r_1) = \theta_a(r_2)$, then $ar_1 = ar_2$ so $a(r_1 - r_2) = 0$.  This implies $r_1 - r_2 = 0$, since $a \not \mid 0$.  Thus $r_1 = r_2$, and we see that $\theta_a$ is injective.  Since $A$ is finite, $\theta_a$ is then also surjective, so there exists $b \in A$ with $\theta_a(b) = 1$, or $ab = 1$.  We have thus shown that every $a \in A$ has a multiplicative inverse, so $A$ must be a field.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Let $(A, \cdot, 1)$ a finite monoid (http://en.wikipedia.org/wiki/Monoid). Let $a$ in $A$ so that the multiplication on the left map 
$$L_a \colon A \to A\\
b \mapsto a\cdot b$$
is injective. Then $a$ is invertible, that is, there exists $a'$ in $A$ so that
$$a \cdot a'= a'\cdot a = 1$$
Indeed, since the set $A$ is finite the map $L_a$ is also surjective and so there exists
$c$ in $A$ so that 
$$L_a(c) = a \cdot c = 1$$
Consider now the multiplication on the right:
$$R_a\colon A \to A \\
b \mapsto b \cdot a$$
From associativity we get 
$$R_c \circ R_a = R_{ac}$$
But $ac = 1$ and so $R_{ac} = R_1 = \mathbb{1}_A$. Therefore
$$R_c\circ R_a= \mathbb{1}_A$$. 
We conclude that the map $R_a$ is injective and so, again, surjective. So there exists
$d$ in $A$ so that
$$R_a(d) = d \cdot a=1$$
We now have 
$$d\cdot a = a \cdot c = 1$$
From here we get 
$$d = d \cdot 1 = d \cdot (a\cdot c) = (d \cdot a) \cdot c = 1 \cdot c = c$$
Thus $d = c$ is the inverse of $a$.
Now consider the finite ring $(A, +, \cdot, 1)$. Assume $a$ is not a left divisor on the left, that is, there does not exist $b \ne 0$ so that $a\cdot b =0$. 
This means: if $b_1 - b_2 \ne 0$ then $a(b_1- b_2)\ne 0$. Use distributivity and conclude: if $b_1 \ne b_2$ then $a \cdot b_1 \ne a \cdot b_2$. But this means that the map $L_a$ is injective. Now apply the previous result. 
