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

Question

The degree of irreducible polynomials over the reals is either one or two.

Is it possible to prove it without using complex numbers? Or without using fundamental theorem of algebra?

Answer 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.

Answer 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.