Finite dimensional $\mathbb R$-algebras without zero divisors

Question

Let $A$ be a finite dimensional commutative algebra over $\mathbb{R}$ without zero divisors. Prove that $\dim_{\mathbb R} A = 2$.

Because any $a \in A\setminus \{0\}$ is not a zero divisors, thus $ab \neq 0$ for $b \neq 0$ and I have no idea.

Answer 1

Let $A$ be a finite dimensional commutative algebra over $\mathbb{R}$, and $j \in A \setminus \mathbb{R}$. Find the nonzero monic polynomial $f \in \mathbb{R}[x]$ of lowest degree such that $f(j) = 0$. Such a polynomial necessarily exists, because otherwise there would be no finite dependence relation on powers of $j$ and therefore $A$ would be infinite dimensional. What is the degree of $f$? It's not $1$ because then $j \in \mathbb{R}$. So $\deg f \geq 2$. If $f$ is reducible then there is a nontrivial factorisation $gh$ in $\mathbb{R}[x]$ of $f$; in that case, $g(j)h(j)=0$ which means either $g(j) = 0$ or $h(j) = 0$ which implies $f$ isn't of lowest degree. Since $\deg f \geq 2$ and $f$ is irreducible, $f$ must be quadratic (because polynomials over $\mathbb{R}$ of degree greater than 2 are reducible). Completing the square on $f$ produces $(x - a)^2 + b$ where $b > 0$. Let $i = \frac{j-a}{\sqrt{b}}$. $i^2 = -1$. So $\mathbb{C} \subseteq A$. Select any $k \in A$ and find a nonzero monic $\phi \in \mathbb{C}[x]$ of lowest degree such that $\phi(k) = 0$; if $\deg \phi > 1$ then we have a nontrivial factorisation of $\phi$ as $\gamma \cdot \eta$ in $\mathbb{C}[x]$ which implies either $\gamma(k) = 0$ or $\eta(k) = 0$ which means $\phi$ is not of lowest degree. So $\phi$ has degree $1$ which means that $k \in \mathbb{C}$. Since this is true for all $k \in A$, we conclude $A \cong \mathbb{C}$.

By the way, the nonzero monic polynomial of lowest degree that sends $x$ to $0$ is called the minimal polynomial of $x$.

Answer 2

Let $a\in A\setminus\mathbb{R}$.

The smallest subring of $A$ containing both $\mathbb{R}$ and $a$ is $\mathbb{R}[a]$, that is by hypothesis a commutative integral domain. Having $0<\dim_{\mathbb{R}}\mathbb{R}[a]<+\infty$, it must be a field, and precisely $\mathbb{R}[a]\cong\mathbb{C}$.

So, without loss of generality we can assume that $a^2+1=0$ and call $\mathbb{C}:=\mathbb{R}[a]\subseteq A$.

Now, let $b\in A$.

Since $A$ is commutative, the smallest subring of $A$ containing both $\mathbb{C}$ and $b$ is indeed $\mathbb{C}[b]$. Again, it has finite dimension as a $\mathbb{R}$-vector space, hence it has finite dimension as a $\mathbb{C}$-vector space. Therefore, since it is an integral domain, it's an algebraic field-exstension of $\mathbb{C}$. But $\mathbb{C}=\overline{\mathbb{C}}$, hence $b\in\mathbb{C}$.

Which proves $A\subseteq\mathbb{R}[a]$. Q.E.D.

Answer 3

Since $A$ is a field (as it was already noticed) we have a finite field extension $\mathbb R\subset A$. By the Primitive Element Theorem $A=\mathbb R(a)$ for some $a\in A$. The degree of our extension is nothing but the degree of the minimal polynomial of $a$ over $\mathbb R$. But we know that the irreducible polynomials over $\mathbb R$ have degree $\le2$.