Show that the ideal $I=\left\langle x_1^2+1,x_2,...,x_n\right\rangle$ is maximal in $\mathbb{R}[x_1,...,x_n]$.

Question

This is an exercise in "Ideals, varieties, and algorithms" by Cox et al.

It first asks to show that $I=\left\langle x^2+1\right\rangle$ is maximal in $\mathbb{R}[x]$. I can show it because it is a PID. Any ideal $J$ containing $I$ can be written as $\left\langle f \right\rangle$ and $f$ must divide $x^2+1$ hence is a nonzero constant.

Then it asks to give an example of a maximal ideal $I$ in $\mathbb{R}[x_1,...,x_n]$ for which $V(I)=\emptyset$. The hint is $\left\langle x_1^2+1,x_2,...,x_n\right\rangle$.

It seems pretty straightforward, but since it is not a PID, I cannot use one function to show contradiction. It is hard to write the generator of an ideal $J$ containing $I$.

Since the book covers quotients of polynomial rings and ring homomorphisms in the next chapter, I think there should be more elementary ways to solve it.

Any help will be appreciated!

Answer 1

Let $f \in \mathbb{R}[x_1,\dots,x_n]$. We want to prove that the ideal generated by $I$ and $f$ is either $I$ or the whole ring.

View $\mathbb{R}[x_1,\dots,x_n]=\mathbb{R}[x_1][x_2,\dots,x_n]$ and write $f(x_1,\dots,x_n)=g(x_1)+h(x_1,\dots,x_n)$, with $h \in I$.

Now write $g(x_1)=(x_1^2+1)q(x_1)+r(x_1)$, with $r=0$ or $r$ has degree 1.

If $r=0$, then $f \in I$.

If $r$ has degree 1, then $r$ and $x_1^2+1$ are coprime, because $x_1^2+1$ is irreducible in $\mathbb{R}[x_1]$. But then some combination of $r$ and $x_1^2+1$ gives $1$ and the ideal generated by $I$ and $f$ is the whole ring.

Answer 2

An alternative way to see that $I$ is maximial is to note that $\mathbb R[x_1,\ldots,x_n]\to \mathbb C$ that sends $x_1\mapsto i$ and $x_k\mapsto 0$ for $k>1$ has $I$ as kernel.

We see that $V(I)=\emptyset$ for if $(a_1,\ldots,a_n)\in\mathbb R^n$ is a point with $f(a)=0$ for all $f\in I$, then specifically $a_1^2+1=0$, which is absurd.