How do you show that the degree of an irreducible polynomial over the reals is either one or two?

Solution 1:

1) If you know that every irreducible polynomial over $\mathbb R$ has degree $1$ or $2$, you immediately conclude that $\mathbb C$ is algebraically closed:

Else there would exist a simple algebraic extension $\mathbb C\subsetneq K=\mathbb C(a)$ with $[K/\mathbb C]=\operatorname {deg}_\mathbb C a=d\gt 1$.
Then $K=\mathbb C(a)=\mathbb R(i,a)=\mathbb R(b)$ for some $b\in K$ by the primitive element theorem
But then the minimal polynomial $f(X)\in \mathbb R[X]$ of $b$ over $\mathbb R$ would be irreducible over $\mathbb R$ and have degree $\operatorname {deg} f(X)=2d\gt 2$, a contradiction to our hypothesis.

2) That said it is possible to prove that every irreducible polynomial over $\mathbb R$ has degree $1$ or $2$ without using the Fundamental Theorem of Algebra for $\mathbb C$.
The method is due to Lagrange and is described in Samuel's Algebraic Theory of Numbers, pages 44-45.
The method consists in inducting on the largest power $r$ of $2$ dividing the degree $d=2^rl$ ($l$ odd) of an irreducible real polynomial, the result being clear for $r=0$ i.e. for odd $n$.
The proof (highly non trivial) proceeds by a clever application of Viète's formulas expressing the coefficients of a polynomial as symmetric functions of the roots of that polynomial.

3) Another real methods proof uses Galois theory and Sylow $2$-groups.
It can be found in Fine-Rosenberg's Theorem 7.6.1
That elementary and pedagogical book is entirely devoted to all kinds of proofs of the Fundamental Theorem of Algebra.

Solution 2:

Yes, it is possible. Here's a proof, taken from my article Another Proof of the Fundamental Theorem of Algebra (American Mathematical Monthly, vol. 112(1), 2005, pp. 76–78).

It is enough to prove that if $\times\colon\Bbb R^n\times\Bbb R^n\longrightarrow\Bbb R^n$ is a bilinear map such that $(\Bbb R^n,+,\times)$ is a field, then $n=2$. That's so because if $p(x)\in\Bbb R[x]$ is an irreducible polynomial with degree $n$, then $\Bbb R[x]/\langle p(x)\rangle$ is a field extension of $\Bbb R$ whose dimension is $n$ and from it we can induce a field structure on $\Bbb R^n$ for which the addition is the usual one.

So, assume that $n>1$ and that you have defined a product $(x,y)\mapsto x\cdot y$ such that $(\Bbb R^n,+,\cdot)$ is a field. Take any norm $\|\cdot\|$ on $\Bbb R^n$ and define a new norm $|\cdot|$ as follows:$$|x|=\sup_{\|y\|\leqslant1}\|x\cdot y\|.$$Then $|1|=1$ and $(\forall x,y\in\Bbb R^n):|x\cdot y|\leqslant|x||y|$. The series$$\sum_{n=0}^{+\infty}\frac{x^n}{n!}\quad\text{and}\quad\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{(x-1)^n}n$$are both absolutely and locally uniformly convergent with respect to this norm, the first one in $\mathbb{R}^n$ and the second one in $\{x\in \mathbb{R}^n\mid|x-1|<1\}$. Their sums are denoted by $\exp(x)$ and $\log(x)$, respectively. Since the product is commutative, it is easy to prove that$$(\forall x,y\in\Bbb R^n):\exp(x+y)=\exp(x)\cdot\exp(y).$$Furthermore, we never have $\exp(x)=0$, because\begin{align}\exp(x)\cdot\exp(-x)&=\exp(x-x)\\&=\exp(0)\\&=1.\end{align}We have thus defined a continuous group homomorphism $\exp:(\mathbb{R}^n,+)\longrightarrow(\mathbb{R}^n\setminus\{0\},\cdot)$.

It can be proved, just as it is in the case of matrices, that$$\exp\bigl(\log(x)\bigr)=x\quad(x\in\mathbb{R}^n,\ |x-1|<1)\tag1$$and$$\log\bigl(\exp(x)\bigr)=x.\tag2$$for any $x$ in $\Bbb{R}^n$ such that $|\exp(x)-1|<1$.

It follows from $(1)$ that, if $V$ is a neighborhood of $0$, then $\exp(V)$ is a neighborhood of $1$. Therefore, since $\exp$ is also a group homomorphism, it is an open mapping. It can be deduced from this fact that $\exp$ is surjective. Indeed, if $G=\exp(\mathbb{R}^n)$, then $G$ is an open subgroup of $(\mathbb{R}^n\setminus\{0\},\cdot)$, and if $x$ belongs to $(\mathbb{R}^n\setminus\{0\})\setminus G$, then $$G\cdot x\subset\bigl(\mathbb{R}^n\setminus\{0\}\bigr)\setminus G.$$ Accordingly, the complement of $G$ in $\mathbb{R}^n\setminus\{0\}$ is also an open set. Therefore, since $\Bbb{R}^n\setminus\{0\}$ is connected (this is where $n>1$ is used), the complement of $G$ must be empty. In other words, $\exp(\mathbb{R}^n)=\mathbb{R}^n\setminus\{0\}$.

It is a consequence of $(2)$ that $\ker(\exp)$ is discrete, and it is well known that, unless $\ker(\exp)=\{0\}$, this implies the existence of linearly independent vectors $v_1,\ldots,v_m$ in $\Bbb{R}^n$ ($m\geqslant1$) such that $\ker(\exp)=\bigoplus_{k=1}^m\Bbb{Z}v_k$. A second application of the fact that $\exp$ is an open mapping shows that it induces a homeomorphism from $\Bbb{R}^n/\ker(\exp)$ (which is homeomorphic to $(S^1)^m\times\Bbb{R}^{n-m}$) onto $\Bbb{R}^n\setminus\{0\}$. But if $n>2$, the space $\mathbb{R}^n\setminus\{0\}$ would be simply connected, whereas $(S^1)^m\times\Bbb{R}^{n-m}$ is not simply connected when $1\leqslant m\leqslant n$. To avoid a contradiction, it would have to be the case that $\ker(\exp)=\{0\}$. Therefore, $\Bbb{R}^n\setminus\{0\}$ would be homeomorphic to $\Bbb{R}^n$. However, this is impossible. This follows from the fact that in $\Bbb{R}^n$ every compact set $K$ is a subset of some other compact set whose complement is connected, whereas in $\Bbb{R}^n\setminus\{0\}$ this is not true (consider, for instance, $K=S^{n-1}$, the unit sphere in $\Bbb{R}^n$). Therefore, $n=2$ and the theorem is proved.