Is $\langle x_{1}\cdots x_{n}-1 \rangle$ a prime ideal in $\mathbb{C}[x_{1},\dots, x_{n}]$?

Question

I've heard from Google that the algebraic torus, the zero locus of $x_{1}\cdots x_{n}-1=0$, is an affine variety, which means $x_{1}\cdots x_{n}-1$ is an irreducible polynomial, which means $\langle x_{1}\cdots x_{n}-1 \rangle$ is a prime. But I still don't know how I can prove that it is prime ideal. Could anyone give me a hint for approaching this?

Answer 1

Here is a hint (and if you cannot complete it I may come back and do so):

1. If you are not satisfied with $z_1-1$ is irreducible for induction first step, prove that $z_1z_2-1$ is irreducible
2. The induction hypothesis is that for $k\ge 1$ we assume that $z_1\cdots z_k -1$ is irreducible
3. Consider now $f=z_1\cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,h\not\in\mathbb C$.
4. Without loss of generality we can assume that $g(z_1,\dots,z_k,1)=1$, you can then show (use the induction hypothesis) that $$g(z_1,z_2,\dots,z_{k-1},1,z_{k+1})=g(z_1,\dots,1,z_k,z_{k+1})=\dots=g(1,z_2,\dots,z_k)=1$$
5. Can you now show that $g$ is $1$?

Answer 2

Here is a slick way to deal with things that doesn't require any computation. We want to show that the ideal $I := \langle X_{1}X_{2}\cdots X_{n} - 1 \rangle$ is a prime ideal of $\mathbb{C}[X_{1}, \ldots, X_{n}]$. Instead, put $R = \mathbb{C}[X_{1}, \ldots, X_{n-1}]$, and let $f(X_{1}, \ldots, X_{n-1}) = X_{1}\cdots X_{n-1} \in R$. Via the isomorphism $R[X_{n}] \cong \mathbb{C}[X_{1}, \ldots, X_{n}]$, we can then view the ideal $I$ as $\langle fX_{n} - 1 \rangle$ in $R[X_{n}]$.

Now, it is well-known that for any commutative ring $A$ and any element $f \in A$, the localization $A_{f}$ is isomorphic to $A[X]/\langle fX - 1 \rangle$. (One way to see this is that both rings satisfy the universal property of localization at the multiplicative subset $S = \{1, f, f^{2}, \ldots\}$; this is done here. Another beautiful, explicit approach is given here.) Hence, we have that $\mathbb{C}[X_{1}, \ldots, X_{n}]/I \cong R[X_{n}]/\langle fX_{n}-1\rangle \cong R_{f}$. Since $R$ is a domain, any nonzero localization of $R$ is also a domain, whence $I$ is prime, as desired.

Incidentally, thinking about the geometry of this variety (at least the closed points) naturally leads one to this approach, I think. If we think about the points of this variety as solutions $(a_{1}, \ldots, a_{n}) \in \mathbb{C}^{n}$ to the equation $x_{1}\cdots x_{n} = 1$, then it is clear that this is equivalent to requiring that $a_{1} \cdots a_{n-1} \in \mathbb{C}^{\times}$, and $a_{n} = \frac{1}{a_{1}\cdots a_{n-1}}$.