Factorise $x^5-x$ in $\mathbb{C}[x]$ and $\mathbb{R}[x]$

Question

I would like general feedback on my solution to this exercise.

Exercise

Factorise $x^5-x$ in $\mathbb{C}[x]$ and $\mathbb{R}[x]$.

Solution

In $\mathbb{C}[x]$ we have \begin{align*} &\phantom{=}\,\,\,x^5-x\\ &=x(x^4-1)\\ &=x(x^2-1)(x^2+1)\\ &=x(x-1)(x+1)(x^2+1)\\ &=x(x-1)(x+1)(x-i)(x+i) \end{align*}

which is a factorisation into irreducibles because the factors are all linear.

In $\mathbb{R}[x]$ we have, as before, $x^5-x = x(x-1)(x+1)(x^2+1)$.

Assume $x^2+1$ has a root $a\in\mathbb{R}$.

Then, $a\in\mathbb{C}$.

But $(x^2+1)=(x-i)(x+i)$ in $\mathbb{C}[x]$, which has roots $\pm i$. So by the uniqueness of polynomial factorisation, $a\in\mathbb{R}$ is impossible.

So a contradiction is reached, and $x^2+1$ does not have a root $a\in\mathbb{R}$.

So $x^2+1$ is irreducible in $\mathbb{R}[x]$.

Because the rest of the factors are linear terms, $x^5-x = x(x-1)(x+1)(x^2+1)$ is then a factorisation into irreducibles.

Answer 1

Your solution is excellent.

Just for thought, there are some other ways to show a general quadratic is irreducible over $\mathbb{R}$ when we don't need to also know the full factorization over $\mathbb{C}$.

• Since $x^2+1>0$ for every real $x$, it cannot have any real roots.
• More generally, we can complete the square: $3x^2 -18x + 32 = 3(x-3)^2 + 5 > 0$ for every $x$, so $3x^2-18x+32$ has no real roots.
• Check the discriminant $D=b^2-4ac$ in the quadratic formula: For $x^2+1$, $D=-4$. For $3x^2-18x+32$, $D=-60$. When the discriminant is negative, there are no real roots.