Establish a bijective correspondence between the maximal ideals of $\Bbb R[x]$ and points in the upper half plane.
The maximal ideals of $\Bbb R[x]$ are the principal ideals $\langle p\rangle$ where $p$ is a monic irreducible polynomial. I found out that $p$ must be of degree $1$ or $2$ with the condition that the discriminant is negative. I know that these two cases give me irreducible polynomials over $\Bbb R$, but why cannot $p$ be of degree $>2$?
If $p$ is of degree say $3$, then $p$ would have $3$ roots, but if all of the roots are complex, then $p$ would still be irreducible over $\Bbb R$? Why is this case not possible? I tried to find a result which would state that a degree $n$ polynomial over $\Bbb R$ would have always at least one real root, but I couldn't find any.
Best Answer
To see that an irreducible polynomial $p\in\mathbb{R}[X]$ can only have degree $1$ or $2$ proceed as follows.
The Fundamental Theorem of Algebra says that $p$ has at least one root in $\mathbb{C}[X]$, say $a+ib$, so $p(a+ib)=0$. Taking complex conjugates we have that $\overline{p(a+ib)}=0$.
For all complex numbers we have that $\overline{z_1 z_2}=\overline{z_1}\overline{ z_2}$ and $\overline{z_1 + z_2}=\overline{z_1}+\overline{ z_2}$; and for real $z$ we have $\overline{z}=z$. Hence we have that $p(\overline{a+ib})=\overline{p(a+ib)}=0$.
So consider the two cases $b=0$ and $b\ne 0$.
In the first case we use the Remainder Theorem in $\mathbb{R}[X]$ to get that $p(X)=(X-a)q(X)$ with $q\in\mathbb{R}[X]$. As $p$ is irreducible in $\mathbb{R}[X]$ we have that $p(X)=c(X-a)$ is of degree $1$.
In the second case we use the Remainder Theorem in $\mathbb{C}[X]$ twice to get that $p(X)=(X^2-2aX+a^2+b^2)q(x)$ with $q\in\mathbb{C}[X]$. However since $p(X)=\overline{p(X)}$ and $\overline{(X^2-2aX+a^2+b^2)}=(X^2-2aX+a^2+b^2)$ we must have that $\overline{q(X)}=q(X)$, and so actually $q\in\mathbb{R}[X]$. However $p$ is irreducible in $\mathbb{R}[X]$ so we must have that $p(X)=c(X^2-2aX+a^2+b^2)$ is of degree $2$.