Prove that any non-zero-divisor of a finite dimensional algebra has an inverse Let $A$ be a finite dimensional algebra. Prove that an element of $A$ is invertible iff it is not a zero divisor.
Let $a$ be an invertible element, then there exists an element $b$ such that $ab=1$ and assyme that $a$ is a zero divisor, then there exists an element $c  \neq 0$ such that $ac=0$ and i don't know.
 A: If $ab=1$ and $ac=0$ then $ab-ac=1$ which means $a(b-c)=1$. So $b-c=b$ (by the uniqueness of inverses) which means that $c=0$, contradicting the fact that $c\neq 0$.
Another short proof for this direction. Suppose $a$ is invertible and let $f\colon A\to A$ be given by $f(x)=axa^{-1}$ which is an algebra homomorphism. The map $f$ is injective because if $f(x)=f(y)$ then $axa^{-1}=aya^{-1}\Rightarrow x=y$. Suppose there exists a $c\in A$ such that $ac=0$. Then $f(c)=aca^{-1}=0a^{-1}=0$ so by injectivity $c=0$ and hence $a$ cannot be a zero-divisor. 
A: Here is another approach.  Let $k$ denote the base field.
The map $T_a : x\mapsto ax$ is a linear transformation $A\to A$ of finite-dimensional vector spaces over $k$ of the same dimension.
By the Rank-nullity theorem, $T_a$ is injective if and only if it is surjective.
But $T_a$ is injective if and only if $a$ is a non-zero-divisor, and surjective if and only if $a$ is a unit.

For a possibly non-commutative algebra, we should consider two linear maps, $L_a$ and $R_a$, for left- and right-multiplication respectively.
Since left invertibility is equivalent to right invertibility, we can now say: $a$ is not a left zero divisor iff $R_a$ is injective iff $R_a$ is surjective iff $a$ is a unit iff $L_a$ is surjective iff $L_a$ is injective iff $a$ is not a right zero divisor.
A: Daniel's answer takes care of one direction of implication.  Note that the fact that the algebra is finite-dimensional is not used.
For the other direction: suppose that $a$ is not invertible.  Note that there must exist a minimal $n$ such that the elements $1,a,a^2,\dots,a^n$ are not linearly independent.  Thus, we have
$$
\sum_{k=0}^n c_k a^k = 
\sum_{k=1}^n c_k a^k = 0 \implies 
a\left(\sum_{k=1}^n c_k a^{k-1}\right) = 0
$$
Note that, in the above, we use the fact that $c_0$ must be $0$ if $a$ is not invertible.  Try to figure out why.
Here is an alternate proof of the other direction: suppose that $a$ is a zero-divisor, and that $c \neq 0$ is such that $ac = 0$.  Suppose (for contradiction) that $a$ has a left inverse.  That is, there is a $b$ such that $ba = 1$. We note that $bac = (ba)c = c$, but $bac = b(ac) = 0$.  So, $c = 0$, which is a contradiction.
A: In any ring, invertible elements (units) cannot be zero divisors; for let $a$ be invertible in the ring $R$; then we have $b \in R$ such that $ab = ba = 1$.  If $a$ is a zero divisor, then we have either $ac = 0$ or $ca = 0$ for some $0 \ne c \in R$.  But $ac = 0$ implies
$c = 1c = (ba)c = b(ac) =0, \tag{1}$
likewise, $ca = 0$ also implies $c = 0$; these contradictions rule out the possibility that $a$ is a zero divisor.
To see that non-zero divisors are invertible for $a \in A$, an algebra of finite dimension over some field $\Bbb F$, recall that for such $a$ the sequence $1, a, a^2, \ldots, a^i, \ldots$ must exhibit a linear dependence since $\dim A < \infty$.  Thus for some smallest $n$ there is a set of $ f_i \in \Bbb F$, $0 \le i \le n$, not all $f_i$ zero,  with
$\sum_0^n f_i a^i = 0; \tag{2}$
here we have $f_n \ne 0$, by virtue of the minimality of $n$; note we may also assume $n \ge 2$, since otherwise (2) asserts that $a \in \Bbb F$ and the question trivializes.  Given a relation as (2) amongst the powers of $a$, we may write
$\sum_1^n f_i a^i = -f_0 \tag{3}$
or
$a\sum_1^n  f_i a^{i -1}= - f_0. \tag{4}$
We note that we cannot have
$\sum_1^n f_i a^{i - 1} = 0 \tag{5}$
by the minimality of $n$; thus we cannot have $f_0 = 0$, for then (4) becomes
$a\sum_1^n f_i a^{i - 1} = 0, \tag{6}$
affirming that $a$ is a zero-divisor.  But with $0 \ne f_0 \in \Bbb F$ we may write (4) in the form
$a (\sum_1^n (-f_0^{-1})f_i a^{i - 1}) = (\sum_1^n (-f_0^{-1})f_i a^{i - 1}) a = 1, \tag{7}$
showing that $a^{-1}$ is in fact given by the polynomial expression
$a^{-1} = \sum_0^n (-f_0^{-1})f_i a^i; \tag{8}$
thus non-zero divisors in algebras $A$ of finite dimension over their fields $\Bbb F$ are invertible.  QED.
