Maximal ideal and not algebraically closed field This is a question concerning maximal ideals in a polynomial ring over a non-algebraically closed field k.
First is the example inspired for this question: as a standard exercise, it is easy to show that $\langle x^2+1\rangle$ is a maximal ideal in $\mathbb{R}[x]$ (Briefly speaking, let $J$ be an ideal s.t. $I\subset J$ and take $f\in J\setminus I$, divide $f$ by $x^2+1$ to obtain $f(x)=q(x)(x^2+1)+(ax+b)$, then one can show that $a^2+b^2\in J$ (see here, for example), so $J=\mathbb{R}[x]$)
Next, consider the ideal $\langle x_1^2+1, x_2,\cdots ,x_n\rangle \subseteq \mathbb{R}[x_1,\cdots ,x_n]$. This is a maximal ideal (use the exact same argument above)
Now, we can generalize things. If $k$ is non-algebraically closed field, we should be able to construct a maximal ideal in $k[x_1,\cdots ,x_n]$ using the logic above. Namely, pick $f\in k[x_1]$ s.t. it has no root in $k$, then consider the ideal $\langle f,x_2,\cdots ,x_n\rangle$. Our inspiration tells us that this ideal should be maximal. However, I have a hard time proving it.
As before, let $J$ be an ideal s.t. $I\subset J$ and take $g\in J\setminus I$. Using the multivariable division algorithm to divide $g$ by $(f,x_1,\cdots ,x_n)$, we obtain $g=q_1f+q_2x_2+\cdots +q_nx_n+r$ with $r\in k[x_1]$. A simple observation shows that $r\in J$. But the biggest problem is I don't see any nice trick to use to show that $J$ contain some nonzero constant (and hence $J=k[x_1,\cdots ,x_n]$)
Any idea?

The Hilbert Nullstellensatz for non-algebraically closed fields describes
all maximal ideals $I$ of $R=k[x_1,\ldots,x_n]$. It states that $R/I$ is a finite
field extension of $k$. This means that $I$ is the kernel of a
$k$-algebra homomorphism $\phi:R\to k^{\text{alg}}$. Such a map is given by
$\phi(f(x_1,\ldots,x_n))=f(a_1,\ldots,a_n)$ with the $a_i\in k^{\text{alg}}$.
Then $I$ is the intersection of $R$ with the ideal of $k^{\text{alg}}[x_1,\ldots,x_n]$ generated by the $x_i-a_i$.
Your example is the case $(a_1,\ldots,a_n)=(a,0,\ldots,0)$. Then $I=\langle
f(x_1),x_2,\ldots,x_n\rangle$ where $f$ is the minimum polynomial of $a$.
