Show that a finite dimensional algebra D with identity over a skewfield F is a semifield if and only if it has no zero divisors.

Question

I'm struggling with a proof of the next lemma.

Show that a finite dimensional algebra $D$ with identity over a skewfield $F$ is a semifield if and only if it has no zero divisors.

EDIT:

Actualy I have skewfield $F$ with centre $Z$ and $\theta$ which is anti-automorphism of $F$ of finite order $m$ and $a, b$ elements of $F$, all such that: $F$ has finite dimension over $Z$ and $x x^{\theta}$ is in $Z$ for all $x$ in $F$a and that $a=x^{\theta +1} + xb$ has no solution for $x$ in $F$.

Now I have two-dimensional vector space $D$ over $F$ with basis elements 1 and $\lambda$, where we use multiplication $(x + \lambda y)(z + \lambda t)=(xz + aty^{\theta}) + \lambda (zy + x^{\theta} t + y^{\theta}bt)$ so that $D$ becomes semifield. So for semifield I would have to check that for or every $x + \lambda y$ and every nonzero $p + \lambda q$ in $D$, there exist unique $z + \lambda t$ and $u + \lambda v$ in $D$ for which $(x + \lambda y)(z + \lambda t) = p + \lambda q$ and $(u + \lambda v)(x + \lambda y) = p + \lambda q$.

But in my book they say that because of the lemma above we only have to show that the only solutions for $(x + \lambda y)(z + \lambda t) = 0$ are $(x + \lambda y) = 0$ or $(z + \lambda t) = 0$.

Answer 1

According to your expanded description $D$ is a finite-dimensional algebra over a (commutative) field $Z$.

I claim that for $a \in D$ the following two statements are equivalent:

1. $a$ is not a left zero-divisor (i.e. there exists no $b \in D \setminus \{0\}$ with $ab=0$).
2. $a$ is left invertible (i.e., there exists $c \in D$ with $ca=1$; or equivalently, for all $y \in D$, there exist $x \in D$ such that $xa=y$).

Proof: Consider the map $\mu_a\colon D \to D$, $x \mapsto ax$. This is a $Z$-linear endomorphism of $D$. Since $D$ is a finite-dimensional $Z$-vector space, $\mu_a$ is injective if and only if it is surjective. Now note that $\mu_a$ is injective if and only if statement 1 holds. Similarly, $\mu_a$ is surjective if and only if statement 2 holds.

Now your lemma follows easily by what I just proved and the symmetric statement where 'left' is replaced by 'right'.

Remark. The fact that every non zero-divisor is a unit holds in more general settings. For instance, it is true for Artinian rings. This includes as a special case finite-dimensional algebras over a field, but also finite rings. It is also true for von Neumann regular rings.