The Fundamental Theorem of Algebra: Examples and Counterexamples

By [L1], the polynomial $f$ has a real root $r$. Since $r^3=2>0$, [L3] gives $r>0$. The identity $x^3-r^3=(x-r)(x^2+rx+r^2)$ therefore yields $x^3-2=(x-r)(x^2+rx+r^2)$ over $\mathbb{R}$.

If $x^2+rx+r^2=0$, then $(2x+r{)}^2=4x^2+4rx+r^2=-3r^2,$ which is impossible in $\mathbb{R}$ because the left-hand side is nonnegative while the right-hand side is negative. So the quadratic factor is irreducible over $\mathbb{R}$.

By [L2], let $\sqrt{3}>0$ be the real square root of $3$. Solving the quadratic factor from step 1.1 gives the two nonreal roots $\frac{-r\pm i\sqrt{3}r}{2}\mathrm{.}$ Substituting them back into the factorization from step 1.1 gives the displayed complete factorization over $\mathbb{C}$.

Thus $x^3-2$ has one real linear factor and one irreducible quadratic factor over $\mathbb{R}$, but it splits completely into three linear factors over $\mathbb{C}$.

Using $(\sqrt{2}{)}^2=2$ from [L1], direct expansion gives $(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)=x^4+1.$

The quadratic $x^2+\sqrt{2}x+1$ has discriminant $2-4=-2<0$, so it has no real root; the same is true of $x^2-\sqrt{2}x+1$. Therefore both quadratic factors are irreducible over $\mathbb{R}$.

Their roots in $\mathbb{C}$ are $\frac{-\sqrt{2}\pm i\sqrt{2}}{2}\text{and}\frac{\sqrt{2}\pm i\sqrt{2}}{2},$ the four fourth roots of $-1$. Pairing each nonreal root with its conjugate recovers the two real quadratic factors from step 1.1.

Hence $x^4+1$ is a concrete polynomial that is irreducible over neither $\mathbb{R}$ nor $\mathbb{C}$, but over $\mathbb{R}$ it factors exactly into two irreducible quadratics.

Fact [L1] gives the real-side claim immediately: $x^2+1$ is irreducible over $\mathbb{R}$.

In $\mathbb{C}[x]$ one has $(x-i)(x+i)=x^2+1,$ so $f$ has the two complex roots $i$ and $-i$. This agrees with the general splitting statement [L2].

Thus $x^2+1$ is the standard smallest-degree polynomial that is irreducible over $\mathbb{R}$ but becomes reducible over $\mathbb{C}$.

Direct evaluation gives $f(1)=1-1-1=-1\text{and}f(2)=32-2-1=29.$

The sign change in step 1.1 is exactly the kind of endpoint behaviour used in the proof of [L1]. Since $f$ has odd degree, [L1] gives a real root.

Thus $x^5-x-1$ is a concrete odd-degree polynomial with a real zero.

The prime $3$ divides the coefficients $-6$ and $3$, does not divide the leading coefficient $1$, and $3^2=9$ does not divide the constant term $3$. So [L1] makes $f$ irreducible over $\mathbb{Q}$.

Let $a:=(6\mathrm{/}5{)}^{1\mathrm{/}4}>1$, which exists by [L2]. For real numbers $x<y$ one has $f(y)-f(x)=(y-x)(y^4+y^{3}x+y^2{x}^2+y{x}^3+x^4-6)\mathrm{.}$ If $a\le x<y$ or $x<y\le -a$, then each of the five degree-four monomials in parentheses is at least $a^4=6\mathrm{/}5$, and at least one is strictly larger than $6\mathrm{/}5$; hence the parenthesis is $>5a^4-6=0$, so $f(y)>f(x)$. If $-a\le x<y\le a$, then each of those monomials is at most $a^4=6\mathrm{/}5$, and they cannot all equal $6\mathrm{/}5$ when $x<y$: equality in the $x^4$ and $y^4$ terms would force $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }=a$, while $x<y$ would then give $(x,y)=(-a,a)$ and hence $y^{3}x=-a^4$. So the parenthesis is $<5a^4-6=0$, and therefore $f(y)<f(x)$. Thus $f$ is increasing on $(-\mathrm{\infty },-a]$, decreasing on $[-a,a]$, and increasing on $[a,\mathrm{\infty })$.

Because $a^4=6\mathrm{/}5<16=2^4$ and $a>0$, one has $a<2$. Also $f(-a)=\frac{24a}{5}+3>0\text{and}f(a)=3-\frac{24a}{5}<0.$ Together with $f(-2)=-17<0$ and $f(2)=23>0$, the continuity from [L3] and the intermediate value theorem [L5] give a root in each of the three intervals $(-2,-a),(-a,a),(a,2)\mathrm{.}$ Step 1.2 shows that $f$ is monotone on each of the three corresponding regions, so there is at most one root in each. Therefore $f$ has exactly three real roots.

By [L6], choose all five roots of $f$ in $\mathbb{C}$ and let $E\subseteq \mathbb{C}$ be the subfield they generate over $\mathbb{Q}$. Fact [L11] makes $E$ a splitting field of $f$ over $\mathbb{Q}$. Step 2.1 shows that exactly three of those roots are real, so the remaining two roots are nonreal. Complex conjugation on $\mathbb{C}$ fixes $\mathbb{Q}$ and preserves the root set, hence it restricts to a $\mathbb{Q}$-automorphism of $E$ that fixes the three real roots and swaps the two nonreal roots. Thus $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ contains a transposition.

The field $\mathbb{Q}$ has characteristic $0$, so [L4] makes it perfect. Step 1.1 shows that $f$ is irreducible over $\mathbb{Q}$, and therefore $f$ is separable by the defining property of a perfect field in [L4]. Fact [L7] now makes $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ transitive on the five roots. With step 3.1, fact [L8] gives $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})=S_5\mathrm{.}$

By [L9], the group $S_5$ is not solvable. If $f$ were solvable by radicals, [L10] would force $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ to be solvable, contradicting step 4.1. Hence $f$ is not solvable by radicals.

Since $f$ is irreducible and has degree $2$, it has no real root.

Thus $f$ is a real polynomial without any real zero.

This single counterexample refutes the claim that every real polynomial has a real root.

By [L2], the polynomial $x^2+1$ has no real root.

Therefore $\mathbb{R}$ fails the defining property in [L1] for algebraic closedness.

So the real numbers are not algebraically closed.

By [L1], the polynomial $x^2+1$ is irreducible over $\mathbb{R}$.

It has degree $2$, and [L2] says that degree $2$ is the exact non-linear possibility for irreducible real polynomials.

Therefore the degree-one claim is false.

Eisenstein at the prime $2$ shows that $q(x)=x^5-2$ is irreducible over $\mathbb{Q}$, so it is an irreducible quintic.

By [L2], choose $s$ with $s^2=5$, choose $\beta$ over $\mathbb{Q}(s)$ with ${\beta }^2=\frac{-5-s}{2},$ and choose $\alpha$ with ${\alpha }^5=2$. Put $u:=\frac{-1+s}{2}\text{and}\zeta :=\frac{u+\beta }{2}\mathrm{.}$ Then $u^2+u-1=0$, ${\zeta }^2-u\zeta +1=0$, and therefore $\zeta +{\zeta }^{-1}=u\mathrm{.}$ Using $u^2+u-1=0$, this implies ${\zeta }^4+{\zeta }^3+{\zeta }^2+\zeta +1=0.$ Since $\zeta \mathrm{\ne }1$, we get ${\zeta }^5=1$, so $\zeta$ is a primitive fifth root of unity. Thus every root of $q$ has the form $\alpha {\zeta }^k$ with $0\le k<5$, and all of them lie in $L:=\mathbb{Q}(s,\beta ,\alpha )\mathrm{.}$

The field tower $\mathbb{Q}\subseteq \mathbb{Q}(s)\subseteq \mathbb{Q}(s,\beta )\subseteq \mathbb{Q}(s,\beta ,\alpha )=L$ adjoins successively a square root of $5$, a square root of $(-5-s)\mathrm{/}2$, and a fifth root of $2$. So $L\mathrm{/}\mathbb{Q}$ is a radical extension. Step 1.2 shows that the splitting field of $q$ is contained in $L$, and [L3] therefore makes $q$ solvable by radicals.

Hence $x^5-2$ is an irreducible quintic over $\mathbb{Q}$ that is solvable by radicals, so the universal claim is false.