Example of prime ideals in $\mathbb{R}[x]$

Question

What are some examples of prime ideals in $\mathbb{R}[x]$? I know $(x^2+1)$ is prime because it only has zeros in $\mathbb{C}$ because $x^2+1=0$ when $x=i$. So $(x^3+1)$ is not prime, $(x^4+1)$ is prime. What about $(x^5+1)$?

Answer 1

$(x^4 + 1)$ isn't a prime ideal in $\Bbb R[x]$, as $(x^2 + \sqrt2x+1)(x^2-\sqrt2x+1) = x^4+1$ is an element of the ideal, but neither of the two factors are. A polynomial doesn't have to have roots to have factorisations.

In fact, any real non-constant polynomial may be factored into a product of quadratic real polynomials and linear real polynomials (a consequence of the funtamental theorem of algebra, plus the fact that $\Bbb C$ has degree $2$ as a field extension over $\Bbb R$, implying that you can't have irreducible polynomials of degree higher than $2$). So the only primes are $(0), (x-a)$ for any $a\in \Bbb R$, and $(x^2+ax+b)$ for any $a, b\in \Bbb R$ with $a^2-4b<0$.

Answer 2

Your argument does not work. The ideal $(x^2 + 1)$ is prime (the quotient is $\mathbb{C}$), but $(x^4+1)$ is not prime. That is because we have $x^4+1 = -(-x^2 + \sqrt{2}x - 1)(x^2 + \sqrt{2}x +1),$ i.e. the polynomial is reducible and leads to zero divisors in the quotient. The root argument only works in degree $2$ and $3$. Now polynomials of odd degree always have a real root, such that you can split them non-trivially if the degree is bigger or equal to $3$ which also makes them reducible then. The ones of even degree are always given by products of real quadratic polynomials, such that they will also be reducible if the degree is bigger or equal than $4$.

To summarize, you will get the zero ideal, ideals generated by linear polynomials and ideals generated by quadratic polynomials not having a real root.