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I believe that this claim may actually be false. If $A$ is a finite-dimensional $K$-algebra then $A$ is a ring with identity together with a ring homomorphism $\varphi: F \rightarrow A$ mapping $1_F$ to $1_A$ such that the subring $\varphi(F)$ is contained in the center of $A$ (Definition, Dummit and Foote p.342).

In order for $A$ to be a division ring, we must first have that $1_A \neq 0_A$. But if $A$ is a ring with $1_A = 0_A$ (so that in fact $A$ is the zero ring) I believe that $A$ could still satisfy the criterion outlined in the problem. The ring homomorphism $\varphi$ can still have $\varphi(1_F) = 1_A = 0_A$, in fact the counterexample I had in mind was to let $\varphi$ be the zero-homomorphism ($\varphi(x) = 0_A$ for all $x \in F$). Since in this instance, the center of $A$ would be just $0_A$ we would also have that the image $\varphi(F) = 0_A \subseteq Z(A)$.

Is there a reason for why we cannot allow $\varphi$ to be the zero-homomorphism? And if we are allowed to consider the zero-homomorphism, how can it be shown that $A$ is not identically zero?

Oderus
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  • Apart from the zero ring issue, if both rings are assumed to be initial and nonzero, can you prove the claim? – Plop Aug 30 '22 at 00:41
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    This is analogous to the fact that a finite integral domain is a field. Consider the map from $A$ to itself given by multiplication by a fixed element. Show this map is injective (use the statement about zero-divisors). Then use that injectivity is equivalent to surjectivity for linear maps on finite dimensional vector spaces. Conclude that surjectivity implies the desired result. – morrowmh Aug 30 '22 at 00:45
  • I have some starting ideas, but not sure I'm on the right track. I thought about having a nonzero $a \in A$ being represented as $a = c_1v_1 + ... + c_nv_n$ for coefficients $c_i \in F$ and $v_i \in A$ since $A$ is a finite-dimensional vector space over $K$, but trying to find the multiplicative inverse of this has left me stumped – Oderus Aug 30 '22 at 00:47
  • But how do we know that $A$ is finite? All I know is that it is a finite-dimensional vector space over the field $F$ – Oderus Aug 30 '22 at 00:49
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    That's why I said "analogous". You don't need $A$ to be finite. What you need is that if $\phi$ is a linear map from a finite-dimensional vector space to itself, then $\phi$ is injective if and only if $\phi$ is surjective. – morrowmh Aug 30 '22 at 00:50
  • Ok, so if I let $\phi_a: A \rightarrow A$ be such that $\phi_a(x) = ax$, then $\ker(\phi) = {x \in F: ax = 0} = {0}$ so $\phi$ is an injective map from $A$ to itself, hence $\phi$ is also surjective, but this just means that $A$ is isomorphic to $A$ as an $F$-vector space, wouldn't I need $A$ to be isomorphic to some field? – Oderus Aug 30 '22 at 00:55
  • The reason the zero ring is not considered is that most definitions of "algebra" include a requirement that $1\neq 0$. Otherwise the subject matter of the question is very frequently asked and has many solutions already on the site. – rschwieb Aug 30 '22 at 11:10
  • And before everyone gets excited about the duplicate choice, do not overlook the accepted answer there which begins (and proves): If is a completely arbitrary field (finite or not), then every finite-dimensional associative -algebra without zero-divisors is a division algebra. – rschwieb Aug 30 '22 at 11:11

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Ignoring the problem with zero rings, we can proceed as follows.

Let $a\in A$ be a nonzero element. To show $A$ is a division ring, we need to show that $a$ is a unit. So consider the $F$-linear map $\phi:A\to A$ given by $r\mapsto ar$. We first show that $\phi$ is injective. To do this, suppose $ar=ar'$ for some $r,r'\in A$. Then $a(r-r')=0$ and since $a\not=0$ and $a$ is not a zero-divisor by assumption, we must have $r-r'=0$. Hence $r=r'$ so $\phi$ is injective. But $A$ is a finite-dimensional vector space, so $\phi$ is surjective (see Linear map $f:V\rightarrow V$ injective $\Longleftrightarrow$ surjective for a reference). Therefore there is some $b\in A$ such that $\phi(b)=1$. But this means $ab=1$ so that $a$ is a unit, and we're done.

morrowmh
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