Dropping the condition in Hilbert's Nullstellensatz

Question

Hilbert Nullstellensatz is not valid for a field which is not algebraically closed.

Nullstellensatz implies that every maximal ideal in $K[x_1,\dots,x_n]$ is of the form $(x_1-a_1,\dots,x_n-a_n)$. So I think I need a counterexample if the field is not algebraically closed, or is there another theorem for those fields ? (see here)

Answer 1

In $\mathbb R[x]$, $(x^2+1)$ is a maximal ideal not of the form $(x-a)$.

Answer 2

Let $I=(x^2+1)\subset \mathbb Q[x]$. Then $I$ is a maximal ideal in $\mathbb Q[x]$ which is not of the form $(x-a)$ for any $a\in \mathbb Q$.