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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?

user26857
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user124697
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  • By the definition of prime ideal, if $ab$ is in the ideal, and $a$ is not in the ideal, you can try to prove $b$ must in the ideal. Although I'm not sure about is this work, but I recommend you to try. – kelvin hong 方 Jan 19 '19 at 03:06
  • @kelvinhong方 Yeah, I tried to do that but it gives me nothing; actually, if it may works, than the assumption that $a$ is not in the ideal say something about $b$. However, since we don't know the given generator is irreducible or not, we cannot say anything. – user124697 Jan 19 '19 at 03:12
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    Oh, I see how to solve it; just assume that it is reducible, then we can do the degree counting to get a contradiction. Thanks! – user124697 Jan 19 '19 at 03:18
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    @user124697 In an integral domain, every prime element is irreducible, but the converse is not true. The converse is true in unique factorization domains. Is $\mathbb{C}[x_1,\cdots,x_n]$ a UFD? (I think it is, because $\mathbb{C}$ is a UFD. But I think it's not very trivial) – stressed out Jan 19 '19 at 03:20
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    @stressedout: yes, it is a result of Gauss that if $R$ is a UFD, then so is $R[X]$. (By induction, this shows that any polynomial ring over a UFD in finitely many variables is also a UFD.) – Alex Wertheim Jan 19 '19 at 03:24
  • @AlexWertheim Thanks for the reference. I will remember it. – stressed out Jan 19 '19 at 03:26
  • @stressedout His intuition is quite true, If $x_1x_2...x_n-1$ is irreducible, then $\langle x_1x_2...x_n-1\rangle$ is a maximal ideal, it follows that every maximal ideal is a prime ideal, but it may not work as well, because a prime ideal need not be a maximal ideal. – kelvin hong 方 Jan 19 '19 at 03:28
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    @kelvinhong方 That's not true. An irreducible element generates a maximal idea amongst "principal ideals" only. Not a maximal ideal of the ring itself in general. A multivariate polynomial ring has many non-principal ideals in general. – stressed out Jan 19 '19 at 03:30
  • @stressedout You're right, I mistakenly use the result of $F[x]$, which is not suitable for $F[x_1,...,x_n]$. – kelvin hong 方 Jan 19 '19 at 03:33
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    To show that $f:=x_1\cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $\mathbb C[x_1\cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1\cdots x_{n-1})(h(x_1\cdots x_{n-1})x_n-k(x_1\cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1\cdots x_{n-1})k(x_1\cdots x_{n-1})=1$$ shows that $g$ is indeed constant. – Pierre-Yves Gaillard Jan 19 '19 at 15:42

2 Answers2

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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 $1$?
quantum
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  • Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},\cdots, 1,z_{k+1}) = \cdots = g(1,z_{2},\cdots, z_{k})=1$? – user124697 Jan 19 '19 at 12:34
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    $f(z_1,\dots,1,z_k)=g(z_1,\dots,1,z_k)h(z_1,\dots,z_k)$ is irreducible so either the left or the right factor is $1$. But the right factor cannot be $1$ because $h(z_1,\dots,z_{k-2},1,1)=\prod_{i=1}^{k-2} z_i -1$ (by the assumption that $g(z_1,\dots,z_k,1)=1$). And you do the same with the other equalities – quantum Jan 19 '19 at 13:05
  • Oh, now I see. Now I am thinking about 5. Give me a little bit time :) – user124697 Jan 19 '19 at 13:17
  • Could you give me more hints on 5? I tried several approaches such as counting leading term, but I cannot get the result; I think the problem is turned out to showing that zariski closure of ${(z_{1}, \cdots, z_{k+1}) : z_{i}=1 \text{ for some }i}$ is the whole affine plane. However, it seems not true since, for example, if $k+1=2$, then ${(x,1)} = V(x-1)$ and ${ (1,y)}= V(y-1)$ thus ${(x,y) : x \text{ or } y=1 } = V((x-1)(y-1)) \neq V(0)$. This implies that $g(x,y)-1 = (xyh(x,y) + xq(x)+yr(y) - 1)(x-1)(y-1)$, and now I'm stucked. Could you let me know how to do that? – user124697 Jan 19 '19 at 14:36
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    If $g\ne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors? – quantum Jan 19 '19 at 15:52
  • Aha! Then, $g$ should be an element of $\langle (z_{1}-1) \cdots (z_{k+1}-1) \rangle$, so $g = h(z_{1}, \cdots, z_{k+1}) (z_{1}-1) \cdots (z_{k+1}-1)$, however, since total degree of $g$ must less than total degree of $f$, which is $k+1$, so $h(z_{1}, \cdots, z_{k+1})$ should be a constant, so $g = c (z_{1}-1) \cdots (z_{k+1}-1)$, but this implies $-1 =f(1, 0, \cdots, 0) = g(1, 0, \cdots, 0)h(1, 0, \cdots,0) = 0 \cdot h(1, 0, \cdots,0) = 0$, contradiction.

    Thank you very much! I really appreciate!

     – user124697 Jan 19 '19 at 18:01
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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 $a_{n} = \frac{1}{a_{1}\cdots a_{n-1}}$.

Alex Wertheim
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  • Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated. – user124697 Jan 21 '19 at 02:42