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Let $F$ be a field containing $\mathbb R$ with the property that $\dim_{\mathbb R}F < \infty.$ Then either $F \cong \mathbb R$ or $F \cong \mathbb C.$

I am trying to prove the above statement. I am not supposed to use Frobenius' theorem as it would just spoil the spirit of the current problem. I am told to show that every $x \in F \setminus \mathbb R$ is a root of some non-zero polynomial in $F[x]$ with leading coefficient $1$ and some other things. But I am not really been able to prove anything so far. Please help. Thanks in advance.

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3 Answers 3

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By the uniqueness of the algebraic closure, we have an embedding $F \hookrightarrow \mathbb C$, hence we have $\mathbb R \subset F \subset \mathbb C$. The result follows from $[\mathbb C:\mathbb R]=2$, because this excludes the existence of proper intermediate fields.

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Let $x\in F\setminus\mathbb{R}1$. Since $F$ is finite-dimensional, say of dimension $N\geq1$, the family of $N+1$ vectors $(1,x,x^2,\ldots,x^N)$ is a dependent family. Hence there exists some real numbers $\alpha_0,\ldots,\alpha_N$ such that $(\alpha_0,\ldots,\alpha_N)\neq(0,\ldots,0)$ and such that $$\alpha_N x^N+\cdots+\alpha_1x+\alpha_01=0.$$

Consider now the associated polynomial $P\in\mathbb{R}[X]$: $$P=\alpha_N X^N+\cdots+\alpha_1X+\alpha_0.$$ Now, this polynomial can be factored as a product of irreducible polynomials in $\mathbb{R}[X]$; and by the Fundamental Theorem of Algebra we know that irreducible polynomials of $\mathbb{R}[X]$ are of degree $1$ or $2$. Without loss of generality we may hence assume that $P$ has the form: $$P=X^2+aX+b.$$ (We can discard the case were the degree of $P$ is $1$, as this would mean that $x\in\mathbb{R}1$.)

Since $P$ is irreducible, we have $a^2/4-b>0$. Now, define the element $j$ of $F$ as $$j=\frac1{\sqrt{a^2/4-b}}x+\frac a{2\sqrt{a^2/4-b}}1.$$ We have: $$j^2=\frac1{a^2/4-b}\left(x^2+ax+\frac{a^2}41\right)=\frac1{a^2/4-b}\left(-b+\frac{a^2}4\right)1=-1$$ since $P(x)=x^2+ax+b1=0$.

Observe that there are exactly two elements of $F$ that have a square equal to $-1$, namely $j$ and $-j$. Indeed, if $u\in F$, then the following factorization holds true: $$u^2+1=(u+j)(u-j)$$ since $F$ is commutative, and since $F$ is an integral domain (since $F$ is a field), $$u^2+1=0\iff u=-j\ \text{or}\ u=j.$$

We now conclude by showing that $F=\operatorname{Span}\{1,j\}=\mathbb{R}1\oplus\mathbb{R}j$. First observe that the family $(1,j)$ is independent since $j\not\in\mathbb{R}1$ since $j^2=-1$. Now, the previous discussion shows that given an element $y\in F\setminus\mathbb{R}1$, there exists an irreducible polynomial $Q\in\mathbb{R}[X]$ of degree $2$, say $Q=X^2+\alpha X+\beta$, such that $Q(y)=0$, and the computation we performed earlier shows that the element $$u=\frac1{\sqrt{\alpha^2/4-b}}y+\frac\alpha{2\sqrt{\alpha^2/4-b}}1\in F$$ has a square equal to $-1$, hence this element is either $j$ or $-j$. Hence $$y=-\frac\alpha2\,1\pm\sqrt{\alpha^2/4-b}\,j$$ and in any case belongs to $\operatorname{Span}\{1,j\}$. Hence $F\subset\operatorname{Span}\{1,j\}$ hence $F=\operatorname{Span}\{1,j\}$. It is now obvious that $F$ is isomorphic (field isomorphism) to $\mathbb{C}$.

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  • $\begingroup$ ok , thanks , but that is only one part , there are other parts ; can you please give a proof or sketch of the full problem ? $\endgroup$
    – user123733
    Mar 21, 2015 at 14:39
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    $\begingroup$ I don't understand how this can answer or even hint to an answer of the question posted. The OP isn't, as far as I can see, seeking any polynomial for some $\;x\in F\setminus\Bbb R\;$, but in fact prove that under the given info we either have $\;F=\Bbb R\;$ or $\;F=\Bbb C\;$ . In fact, proving that any $\;x\in F\setminus\Bbb R\;$ has a quadratic minimal polynomial will do the trick. $\endgroup$
    – Timbuc
    Mar 21, 2015 at 14:42
  • $\begingroup$ @Timbuc I don't want to give the full answer (yet). I only addressed the I am told to show that every $x\in F\setminus\mathbb{R}$ is a root of some non-zero polynomial in $F[x]$ with leading co-efficient $1$ part of the OP. $\endgroup$ Mar 21, 2015 at 14:45
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    $\begingroup$ @gniourf_gniourf That much is true for any fields extension so it doesn't use at all this very particular case. I think the gist of this is inthe comtinuation of the phrase you mention: "...and some other things ", which seems to point towards the hint that if $\;F\neq\Bbb R\;$ then the minimal polynomial of any $\;x\in F\setminus\Bbb R\;$ is a quadratic one. $\endgroup$
    – Timbuc
    Mar 21, 2015 at 14:47
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    $\begingroup$ I am assuming FTA ; can user Timbuc or gniourf_gniourf please now give a solution :-) $\endgroup$
    – user123733
    Mar 21, 2015 at 15:55
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Since $F$ is finite dimensional over $\mathbb{R}$ it is algebraic over $\mathbb{R}$. This is a basic fact about field extensions: if $a\in F$, then $1,a,a^2,\dots,a^n$ are not linearly independent over $\mathbb{R}$, where $n=\dim_{\mathbb{R}}F$. So every element of $F$ is the root of a polynomial with coefficients in $\mathbb{R}$.

If $-1$ is not a square in $F$, we can add a square root $j$ (with $j^2=-1$), so we have $F[j]$ which is finite dimensional over $F$, hence also over $\mathbb{R}$ by the dimension formula. Now $\mathbb{R}[j]$ is isomorphic to $\mathbb{C}$ and we have the chain $\mathbb{R}\subseteq \mathbb{R}[j]\subseteq F[j]$ which tells us that $F[j]=\mathbb{R}[j]$, because $\mathbb{C}$ is algebraically closed. Thus $F=\mathbb{R}$.

Otherwise $-1$ is a square in $F$, so $F$ is algebraic over $\mathbb{R}[j]$ (where $j^2=-1$) which is algebraically closed, being isomorphic to $\mathbb{C}$. Hence $F=\mathbb{R}[j]$ and so $F\cong\mathbb{C}$.

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