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If we see the complex numbers $\mathbb{C}$ as a linear space on field $\mathbb{Q}$, then $\mathbb{C}$ is infinite dimensional over $\mathbb{Q}$, $2$-dimensional over $\mathbb{R}$, $1$-dimensional over $\mathbb{C}$ itself.

My question: Is there a field $F\subset\mathbb{R}$, $\mathbb{R}$ is $2$-dimensional linear space over $F$?

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There is no such field $F$. Suppose otherwise. Choose any $x \in \mathbb{R}-F$, so $\{1,x\}$ is a basis of $\mathbb{R}$ over $F$. Since the degree of the extension equals $2$, there is a polynomial $x^2 + ax + b = 0$ with coefficients $a,b \in F$. By the quadratic formula, $x$ is an $F$-linear combination of $1$ and $\sqrt{y}$ for some $y \in F$ and $y$ must be positive since $x$ is real. It follows that $\{1,\sqrt{y}\}$ is a basis of $\mathbb{R}$ over $F$.

Consider $z = \sqrt[4]{y} \in \mathbb{R}$, $z>0$. We can write $z = a + b \sqrt{y}$ where $a,b \in F$. Solving for $a,b$ in the equation $z^2 = \sqrt{y}$ we obtain $a^4 = -\frac{1}{4} y$. However, a negative number has no fourth roots in $\mathbb{R}$, contradiction.

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No. Assume $F$ is such a field and $a\in\mathbb R\setminus F$. Then $1,a,a^2$ are $F$-linearly dependant, i.e. $a$ is algebraic of degree $2$ over $F$, say $a^2=u+va$ with $u,v\in F$. Let $b=|a-v/2|>0$. Then $b\notin F$, but $b^2=u+v^2/2\in F$. We conclude that $x+yb\mapsto x-yb$ (with $x,y\in F$) is a field automorphism of $\mathbb R$ that leaves $F$ fixed and maps $b\mapsto -b$. Since $b>0$, let $c=\sqrt b$. Then $c=z+wb$ for some $z,w\in F$. Then $(z+wb)^2=b$ and via the automorphism described above, $(z-wb)^2=-b<0$, contradiction.

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