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

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... \

How would be this proof? thanks! :)

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!!!

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. $~~\blacksquare$

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.

• +1 for not assuming that the ring is commutative. One should finish by appealing to Wedderburn's theorem, i.e. a finite division ring is a field. In another direction, if the ring is assumed commutative, but not necessarily with $1$, then nonexistence of zerodivisors implies existence of $1$ - see e.g. here – zcn Oct 1 '14 at 8:29
• @zcn: Thanks! Right, Wedderburn would finish it off. I wanted to point out that for general finite rings an element is either a unit or a zero divisor (on both sides). I will take a look at the reference pointed. – orangeskid Oct 1 '14 at 8:47