How to prove that $\Bbb R[x]/\langle x^5+x-3\rangle$ is not an integral domain?

Question

I have used sage to get

sage: x = PolynomialRing(RealField(), 'x').gen()

sage: f = x^5+x-3

sage: f.factor()

$(x - 1.13299756588507) * (x^2 - 0.950761513339099*x + 1.50221285149526) * (x^2 + 2.08375907922416*x + 1.76262856840348)$

How to use a general trick to prove it without using any tool?

score 4 · Answer 1 · answered Jun 24 '19 at 22:26

4

HINT: Use the fact that every polynomial in $\mathbb R[x]$ is a product of quadratic and linear terms.

answered Jun 24 '19 at 22:26

Stefan4024

36,357

2

Done!! What a fool I am!! :( – Ri-Li Jun 24 '19 at 22:27
4

@Hunter: Depending on your taste (i.e., what you'd like to call elementary), there is an even more elementary argument of the same flavor: every odd-degree polynomial over $\mathbb{R}$ has a real root by the intermediate value theorem. – Alex Wertheim Jun 25 '19 at 07:22

score 3 · Accepted Answer · edited Jun 25 '19 at 05:00

3

It is a finite-dimension algebra over a field, so if it is a domain, it is an algebraic field extension. But there are no field extensions of degree 5 over $\mathbb{R}$ since $\mathbb{C}$ is algebraically closed and degree 2 over $\mathbb{R}$.

edited Jun 25 '19 at 05:00

rschwieb

160,592

answered Jun 24 '19 at 22:26

user10354138

33,887

How to prove that $\Bbb R[x]/\langle x^5+x-3\rangle$ is not an integral domain?

2 Answers2