Showing that a two-dimensional unitary commutative algebra over $\mathbb{R}$ has a certain basis. Let $A$ be a two-dimensional unitary commutative algebra over $\mathbb{R}$ with identity $u$. I have to show that $A$ has a basis $u, v$ such that $v^2=ru$ for some $r\in\mathbb{R}$.
My attempt so far is to write $v=\alpha u + \beta w$ for some $\alpha,\beta\in\mathbb{R}$ where $u, w$ is another basis. Then $$v^2=(\alpha u + \beta w)(\alpha u + \beta w) = \alpha^2u+\beta^2w^2+2\alpha\beta w = \alpha^2u+\beta^2(\gamma u+\delta w) + 2\alpha\beta w  =(\alpha^2+\beta^2\gamma)u+(2\alpha\beta+\beta^2\delta)w,$$ where I have written $w^2$ as $\gamma u+\delta w$ for some $\gamma, \delta\in\mathbb{R}$.
I then let $r=\alpha^2+2\alpha\beta$ and try to conclude (somehow) that $2\alpha\beta=0$.
However, I'm not very confident that this approach is correct or even achieves what is being asked. In particular, I don't think it shows that $u,v$ is a basis.
Any help would be greatly appreciated! Thanks in advance
 A: You're on the right track; just finish it! ;-)
Suppose $\{u,w\}$ is a basis for $A$ over $\mathbb{R}$. If $w^2=ru$, for some $r\in\mathbb{R}$, you are done. Otherwise $w^2=\gamma u+\delta w$, for some $\delta\ne0$. Hence
$$
(u+\beta w)^2=
(1+\beta^2\gamma)u+(2\beta+\beta^2\delta)w
$$
Thus you can take
$$
\beta=-\frac{2}{\delta}
$$
Note that $\{u,u+\beta w\}$ is linearly independent whenever $\beta\ne0$.
A: As the structural map: $\mathbb{R} \rightarrow A$ is injective, I'll consider $\mathbb{R}$ as a subring of $A$.
I'll show an even stronger claim: We can find an element $v \in A \setminus \mathbb{R}$, such that $v^2 \in \mathbb{R}$. From this, your statement follows, as we can take $\{1,v\}$ as basis of $A$ with the desired property. It is a basis as the spanned subspace properly contains $\mathbb{R}$, thus is two-dimensional, i.e. whole $A$.
To see this, let $x \in A \setminus \mathbb{R}$. Then, the set $1,x,x^2$ must be linearly dependent, so we have a non-trivial equation of the form $\lambda_1x^2+\lambda_2x+\lambda_3=0$ Note that $\lambda_1 \neq 0$, as $1,x$ is linearly independent, as the subspace generated by it properly contains $\mathbb{R}$, so must be two-dimensional, i.e. $A$. So after divison by $\lambda_1$, we may assume that $\lambda_1 = 1$.
Now we complete the square and get $0=x^2+\lambda_2x+\lambda_3 = (x+\frac{\lambda_2}{2})^2+\lambda_3-\frac{\lambda_2^2}{4}$
Now take $v = x + \frac{\lambda_2}{2}$, then the last equation shows that $v^2 \in \mathbb{R}$ as claimed.
In fact, after normalizing, one can choose $v$ such that $v^2 \in \{-1,0,1\}$ which one can use to give a complete classification of two-dimensional commutative real algebras.
