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Let $\mathbb{F}$ be a field and $\mathbb{F\not=R}$. I want to show that there does not exist a field extension $\mathbb{F}/\mathbb{R}$ of odd degree. Here $\mathbb{R}$ is the field of real numbers.

My attempt: pick $\alpha\in\mathbb{F}, \alpha\notin\mathbb{R}$. By multiplicative nature of degree extensions, [$\mathbb{F:R}$]=[$\mathbb{F:R(\alpha)}$][$\mathbb{R(\alpha):R}$]. I have been trying to show that [$\mathbb{R(\alpha):R}$] is even to no avail.

If I try to show this by contradiction, then one of [$\mathbb{F:R(\alpha)}$] or [$\mathbb{R(\alpha):R}$] should be odd. But I again I get stuck.

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What degree does the minimal polynomial for $\alpha$ have? Draw a picture of the graph of that polynomial. – user29743 Apr 29 '12 at 4:31
@countinghaus, I do not understand how to draw this graph, I am picking an arbitrary element of the field $F$, and I don't know how to find a minimal polynomial of this element. – Edison Apr 29 '12 at 4:34
@Edison: But you know the polynomial has degree $[\mathbb{R}(\alpha):\mathbb{R}]$, and is monic, so you know what happens to the polynomial for large inputs and for large negative inputs. And since polynomials are continuous... – Arturo Magidin Apr 29 '12 at 4:39
@Edison: Your $\mathbb{F}(\alpha)$'s should be $\mathbb{R}(\alpha)$s. – Arturo Magidin Apr 29 '12 at 4:39
@ArturoMagidin, I just edited my post. Thanks. – Edison Apr 29 '12 at 4:41
up vote 6 down vote accepted

Suppose $F$ has odd degree over $\mathbb{R}$; let $\alpha\in F$. Then, as you note, $$[F:\mathbb{R}] = [F:\mathbb{R}(\alpha)][\mathbb{R}(\alpha):\mathbb{R}].$$ Since $[F:\mathbb{R}]$ is odd, then so is its divisor $[\mathbb{R}(\alpha):\mathbb{R}]$. Therefore, the minimal irreducible polynomial of $\alpha$ over $\mathbb{R}$ has odd degree.

If $p(x) = x^{2n+1}+\cdots + a_1x + a_0$ is the minimal irreducible polynomial for $\alpha$, then note that $\lim\limits_{x\to\infty}p(x) =\infty$ and $\lim\limits_{x\to-\infty}p(x)=-\infty$. By the intermediate value theorem, $p(x)$ has a root $c$. Therefore, by the factor theorem $x-c$ divides $p(x)$. Since $p(x)$ is irreducible and monic, $p(x)=c-x$, so $\alpha=c$. Therefore, $\alpha\in\mathbb{R}$.

Since $\alpha\in\mathbb{R}$ for all $\alpha\in F$, we conclude that $F\subseteq\mathbb{R}$. Hence, $F=\mathbb{R}$. So the only extension of $\mathbb{R}$ of odd degree is $\mathbb{R}$ itself, which has degree $1$.

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this is a very nice proof, thanks! – Edison Apr 29 '12 at 4:59

Here is an application of that formula to a problem similar to yours above:

There is no extension of $\Bbb{R}$ of degree 4.

Suppose we have an extension $F$ of the reals of degree 4. Then such an extension $F$ is $\Bbb{R}[a_1,\ldots a_n]$ where the $a_i \in F$ (why?). Now since $F$ is an extension of $\mathbb{R}$ of degree 4 we are adjoining at most two elements of $F$ to $\Bbb{R}$. This is because if say we adjoin three elements $a,b,c$ to $\Bbb{R}$, without loss of generality we have

$$[\Bbb{R}(a,b,c) : \Bbb{R}] = [\Bbb{R}(a,b,c) : \Bbb{R}(a,b) ] [\Bbb{R}(a,b), \Bbb{R}(a) ] [\Bbb{R}(a):\Bbb{R}] $$

or that we have 3 positive integers multiplying together to give 4. Therefore one of those extensions must have been the trivial extension. Similarly adjoining 4 or more elements does not create any bigger field than adjoining two elements. We now show that assuming $F$ is an extension of $\Bbb{R}$ by adjoining one or two elements of $F$ leads to a contradiction. Suppose $F = \Bbb{R}(a)$. Then this means that $a \in F$ is the root of a real monic irreducible polynomial of degree 4. However this is not possible because every real polynomial of degree 4 is reducible over $\Bbb{R}$ (exercise; use the fundamental theorem of algebra). So the only possibility left is that $F$ is obtained from $\Bbb{R}$ by adjoining two elements, which we call $\alpha $ and $\beta$. Write $$[\Bbb{R}(\alpha,\beta) : \Bbb{R}] = [\Bbb{R}(\alpha,\beta):\Bbb{R}(\alpha)][\Bbb{R}(\alpha):\Bbb{R}].$$

For the right hand side to equal 4 this means that each factor in the right hand side is 2. We cannot have one of them being $1$ because then the other is 4 contradicting the fact that all real polynomials are reducible over $\Bbb{R}$. Now if $[\Bbb{R}(\alpha):\Bbb{R}]$ is 2 then this means that $\alpha$ is a root of a real polynomial of degree 2, so that $\alpha$ is contained in the complex numbers, hence $\Bbb{R}(\alpha) \subset \Bbb{C}$. But then we know that there are no proper fields of finite degree as an extension of $\Bbb{R}$ between $\Bbb{C}$ and $\Bbb{R}$. This forces $\Bbb{R}(\alpha) = \Bbb{C}$.

It follows that the extension $\Bbb{R}(\alpha,\beta)/\Bbb{R}(\alpha)$ is an extension of $\Bbb{C}$ of degree two. However this means that $\beta$ is the root of an irreducible polynomial of degree 2 with complex coefficients. However since $\Bbb{C}$ is algebraically closed the only irreducible polynomials in $\Bbb{C}$ are those of degree 1. This is a contradiction so that the right hand side of equation $(1)$ can never equal 4. It follows that there is no field extension of $\Bbb{R}$ of degree 4.

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Thank you for this neat application! – Edison Apr 29 '12 at 4:51

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