How to find all the prime and maximal ideals of $\mathbb R [x]$?

Question

I know that $\mathbb{R}[x]$ is a PID. I have proven that all the ideals of the form $I=(x)$ are maximals ($\mathbb{R}[x]/(x)$ is isomorph to $\mathbb{R}$ which is a field. As a result I is maximal). I also proved that if $f(x)\in\mathbb{R}[x]$ is ireducable then $\mathbb{R}[x]/f(x)$ is isomorph to $\mathbb{C}$ which is a field, so $<f(x)>$ is maximal. Are those the only maximals ideals? How can I prove it? Are there any other prime ideals that are not maximals?

Answer 1

The zero ideal is prime but not maximal. All the other prime ideals are maximal. Indeed, a nonzero prime ideal is generated by a prime element $f\in\mathbb{R}[x]$. But prime elements in an integral domain are irreducible, and so $f$ must be an irreducible polynomial. So now, if $(f)\subseteq (g)$ for some monic polynomial $g\in\mathbb{R}[x]$ then $g|f$, and since $f$ is irreducible it follows that $g=f$ or $g=1$. Thus $(g)=\mathbb{R}[x]$ or $(g)=(f)$. So indeed $(f)$ is a maximal ideal.

By the way, note that $f$ being irreducible doesn't imply that $\mathbb{R}[x]/(f)$ is isomorphic to $\mathbb{C}$. For example, if $f$ has degree $1$ then $\mathbb{R}[x]/(f)$ is isomorphic to $\mathbb{R}$. If $f$ is irreducible of degree $2$ then indeed $\mathbb{R}[x]/(f)\cong\mathbb{C}$.

Answer 2

Let $I=(f)$ be an ideal of $\mathbb{R}[x]$; if $f$ is a constant non-zero (the ideal $(0)$ is not maximal) polynomial (i.e. it has degree $0$) then it is invertible; so $f$ must have degree $\geq1$. You have proved that, given that $f$ is irreducible, then $I$ is maximal (it is true indeed, but it is not true that the quotient is always isomorphic to $\mathbb{C}$). So, let us suppose this is not the case, and so $f(x)=p_1(x)\ldots p_n(x)$, with $p_i(x)$ non-constant and coprime with $p_j$, $j \neq i$; then the quotient $\mathbb{R}[x]/I \simeq \mathbb{R}[x]$ is not a field, since it can be splitted into a product of rings using the Chinese remainder Theorem for ideals.