# How can I show that $\mathbb R [x]/(x^5+x-3)$ is not an Integral domain? [duplicate]

How can I show that $$\mathbb R [x]/(x^5+x-3)$$ is not an Integral domain?

To prove that it is not an Integral domain at first I have show that $$(x^5+x-3)$$ this ideal is not a prime ideal as $$\mathbb R [x]$$ is a commutative ring with unity. For this I have to show $$a(x)b(x)$$ product of two polynomials belongs to the ideal $$(x^5+x-3)$$ but $$a(x)\ \&\ b(x) \notin (x^5+x-3)$$.

I can't find such $$a(x)\ \&\ b(x)$$?

Is there any other method to do this?

Please suggest some edit.

• You don't really need to show factors, just show that they exist. Let $r$ be any root of $x^5+x-3$, then either $x-r$ is a (real polynomial) factor or $(x-r)(x-\overline{r})$ is one. – conditionalMethod Nov 8 '19 at 10:24
• So, as long as the degree is $\geq 3$ it will factor over $\mathbb{R}$. – conditionalMethod Nov 8 '19 at 10:31

Since $$x^5+x-3$$ is an odd dgree polynomial, it has some real root $$r$$. So, you can write it as $$(x-r)\times q(x)$$ for some polynomial $$q(x)$$ with degree $$4$$ and therefore in $$\mathbb R[x]/(x^5+x-3)$$ you have $$(x-r)\times q(x)=0$$.
Any real polynomial of odd degree has a real root. In particular, $$f=x^5+x-3$$ is not irreducible and hence not prime. Note that $$\Bbb R[X]$$ is a factorial ring. But $$\Bbb R[X]/(f)$$ is an integral domain if and only if $$(f)$$ is a prime ideal.