17

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.

user26857
  • 52,094

3 Answers3

15

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.

MooS
  • 31,390
7

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 $\mathbb{C}$.

gniourf_gniourf
  • 4,196
  • 18
  • 22
  • ok , thanks , but that is only one part , there are other parts ; can you please give a proof or sketch of the full problem ? –  Mar 21 '15 at 14:39
  • 1
    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. – Timbuc Mar 21 '15 at 14:42
  • @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. – gniourf_gniourf Mar 21 '15 at 14:45
  • 1
    @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. – Timbuc Mar 21 '15 at 14:47
  • @Timbuc Not exactly for any field extension. Only the extensions of finite degree. Regarding the remaining part of the proof, I don't know what some other things exactly is, what it involves, etc. (can we use FTA here?). – gniourf_gniourf Mar 21 '15 at 14:53
  • @gniourf_gniourf Yes, of course only algebraic, and among those finite, extensions. I think the easiest way to go is to show that any real polynomial is decomposable as a product of linear and/or quadratic (real) polynomials, unsing the fact that complex roots of real pol's come in conjugate pairs. The bottom fact is that your answer doesn't help at all, as far as I can see, to deduce that it must be $;F=\Bbb R;;or;;F=\Bbb C;$ – Timbuc Mar 21 '15 at 14:56
  • @Timbuc so that's using FTA… is OP ok with the use of FTA? – gniourf_gniourf Mar 21 '15 at 14:57
  • @gniourf_gniourf If the OP can use FTA I can't see how this is a challenging question. I think we can not assume the FTA. – Timbuc Mar 21 '15 at 14:58
  • @Timbuc: how do you show that every real polynomial can be factored as you said, without using FTA? your claim is equivalent to saying that every non-constant polynomial with real coefficients possesses at least one (complex) root; and this is well-known to be equivalent to FTA. – gniourf_gniourf Mar 21 '15 at 15:19
  • @gniourf_gniourf Of course it is not equivalent to FTA! You just have to make clear that if $;p(x)\in\Bbb R[x];$ and $;z\in\Bbb C;$ , then $$p(z)=0\implies p(\overline z)=\overline{p(z)}=\overline0=0$$ I'm talking of the conjugate pairs. Of course, you need to use FTA for the existence of at least one root. – Timbuc Mar 21 '15 at 15:36
  • @Timbuc I'm not talking about the fact that the non-real roots come in conjugate pairs (that's obvious), I'm talking about existence of complex roots: why does such a factorization exist, without using FTA? – gniourf_gniourf Mar 21 '15 at 15:37
  • @gniourf_gniourf Then we're not disagreeing here...:) – Timbuc Mar 21 '15 at 15:38
  • 1
    I am assuming FTA ; can user Timbuc or gniourf_gniourf please now give a solution :-) –  Mar 21 '15 at 15:55
  • @user123733 post edited to add the remaining part of the proof (that relies on the form of irreducible polynomials of $\mathbb{R}[X]$ that needs FTA). – gniourf_gniourf Mar 21 '15 at 16:33
5

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 $F\cong\mathbb{C}$.

egreg
  • 238,574