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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

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2 Answers 2

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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 $\beta\ne0$.

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  • $\begingroup$ Thank you! Maybe a dumb question, but how come you can write $v=u+\beta w$ and not $v = \alpha u+\beta w$? Using $\alpha u+\beta w$ gives $\beta=-\frac{2\alpha}{\delta}$ instead, which does not obviously yield (at least to me) that $\beta\neq 0$. $\endgroup$
    – user424388
    Mar 13, 2017 at 0:11
  • $\begingroup$ @user424388 $u$ must appear, so we can assume its coefficient is $1$, because it only changes the value of $r$ $\endgroup$
    – egreg
    Mar 13, 2017 at 6:28
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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.

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