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












1












$begingroup$


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?






algebraic-geometry commutative-algebra irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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!
    $endgroup$
    – user124697
    Jan 19 at 3:18






  • 1




    $begingroup$
    @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)
    $endgroup$
    – stressed out
    Jan 19 at 3:20








  • 1




    $begingroup$
    @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.)
    $endgroup$
    – Alex Wertheim
    Jan 19 at 3:24






  • 1




    $begingroup$
    @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.
    $endgroup$
    – stressed out
    Jan 19 at 3:30








  • 1




    $begingroup$
    To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
    $endgroup$
    – Pierre-Yves Gaillard
    Jan 19 at 15:42
















1












$begingroup$


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?






algebraic-geometry commutative-algebra irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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!
    $endgroup$
    – user124697
    Jan 19 at 3:18






  • 1




    $begingroup$
    @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)
    $endgroup$
    – stressed out
    Jan 19 at 3:20








  • 1




    $begingroup$
    @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.)
    $endgroup$
    – Alex Wertheim
    Jan 19 at 3:24






  • 1




    $begingroup$
    @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.
    $endgroup$
    – stressed out
    Jan 19 at 3:30








  • 1




    $begingroup$
    To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
    $endgroup$
    – Pierre-Yves Gaillard
    Jan 19 at 15:42














1












1








1





$begingroup$


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?






algebraic-geometry commutative-algebra irreducible-polynomials







share|cite|improve this question











$endgroup$




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?






algebraic-geometry commutative-algebra irreducible-polynomials




algebraic-geometry commutative-algebra irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 20:30









user26857

39.3k124183




39.3k124183










asked Jan 19 at 3:04









user124697user124697

638515




638515








  • 1




    $begingroup$
    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!
    $endgroup$
    – user124697
    Jan 19 at 3:18






  • 1




    $begingroup$
    @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)
    $endgroup$
    – stressed out
    Jan 19 at 3:20








  • 1




    $begingroup$
    @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.)
    $endgroup$
    – Alex Wertheim
    Jan 19 at 3:24






  • 1




    $begingroup$
    @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.
    $endgroup$
    – stressed out
    Jan 19 at 3:30








  • 1




    $begingroup$
    To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
    $endgroup$
    – Pierre-Yves Gaillard
    Jan 19 at 15:42














  • 1




    $begingroup$
    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!
    $endgroup$
    – user124697
    Jan 19 at 3:18






  • 1




    $begingroup$
    @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)
    $endgroup$
    – stressed out
    Jan 19 at 3:20








  • 1




    $begingroup$
    @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.)
    $endgroup$
    – Alex Wertheim
    Jan 19 at 3:24






  • 1




    $begingroup$
    @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.
    $endgroup$
    – stressed out
    Jan 19 at 3:30








  • 1




    $begingroup$
    To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
    $endgroup$
    – Pierre-Yves Gaillard
    Jan 19 at 15:42








1




1




$begingroup$
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!
$endgroup$
– user124697
Jan 19 at 3:18




$begingroup$
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!
$endgroup$
– user124697
Jan 19 at 3:18




1




1




$begingroup$
@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)
$endgroup$
– stressed out
Jan 19 at 3:20






$begingroup$
@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)
$endgroup$
– stressed out
Jan 19 at 3:20






1




1




$begingroup$
@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.)
$endgroup$
– Alex Wertheim
Jan 19 at 3:24




$begingroup$
@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.)
$endgroup$
– Alex Wertheim
Jan 19 at 3:24




1




1




$begingroup$
@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.
$endgroup$
– stressed out
Jan 19 at 3:30






$begingroup$
@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.
$endgroup$
– stressed out
Jan 19 at 3:30






1




1




$begingroup$
To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
$endgroup$
– Pierre-Yves Gaillard
Jan 19 at 15:42




$begingroup$
To show that $f:=x_1cdots x_n-1$ is irreducible, view it as a polynomial in $x_n$ with coefficients in $mathbb C[x_1cdots x_{n-1}]$. Any factorization of $f$ has the form $$f=g(x_1cdots x_{n-1})(h(x_1cdots x_{n-1})x_n-k(x_1cdots x_{n-1})).$$ It suffices to show that $g$ is constant. But the equality $$g(x_1cdots x_{n-1})k(x_1cdots x_{n-1})=1$$ shows that $g$ is indeed constant.
$endgroup$
– Pierre-Yves Gaillard
Jan 19 at 15:42










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $kge 1$ we assume that $z_1cdots z_k -1$ is irreducible

  3. Consider now $f=z_1cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,hnotinmathbb C$.

  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 $1$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
    $endgroup$
    – user124697
    Jan 19 at 12:34






  • 1




    $begingroup$
    $f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
    $endgroup$
    – quantum
    Jan 19 at 13:05












  • $begingroup$
    Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
    $endgroup$
    – user124697
    Jan 19 at 13:17










  • $begingroup$
    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?
    $endgroup$
    – user124697
    Jan 19 at 14:36








  • 1




    $begingroup$
    If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
    $endgroup$
    – quantum
    Jan 19 at 15:52



















1












$begingroup$

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 $mathbb{C}[X_{1}, ldots, X_{n}]/I cong R[X_{n}]/langle fX_{n}-1rangle cong R_{f}$. 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 $a_{n} = frac{1}{a_{1}cdots a_{n-1}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
    $endgroup$
    – user124697
    Jan 21 at 2:42











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $kge 1$ we assume that $z_1cdots z_k -1$ is irreducible

  3. Consider now $f=z_1cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,hnotinmathbb C$.

  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 $1$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
    $endgroup$
    – user124697
    Jan 19 at 12:34






  • 1




    $begingroup$
    $f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
    $endgroup$
    – quantum
    Jan 19 at 13:05












  • $begingroup$
    Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
    $endgroup$
    – user124697
    Jan 19 at 13:17










  • $begingroup$
    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?
    $endgroup$
    – user124697
    Jan 19 at 14:36








  • 1




    $begingroup$
    If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
    $endgroup$
    – quantum
    Jan 19 at 15:52
















1












$begingroup$

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 $kge 1$ we assume that $z_1cdots z_k -1$ is irreducible

  3. Consider now $f=z_1cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,hnotinmathbb C$.

  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 $1$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
    $endgroup$
    – user124697
    Jan 19 at 12:34






  • 1




    $begingroup$
    $f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
    $endgroup$
    – quantum
    Jan 19 at 13:05












  • $begingroup$
    Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
    $endgroup$
    – user124697
    Jan 19 at 13:17










  • $begingroup$
    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?
    $endgroup$
    – user124697
    Jan 19 at 14:36








  • 1




    $begingroup$
    If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
    $endgroup$
    – quantum
    Jan 19 at 15:52














1












1








1





$begingroup$

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 $kge 1$ we assume that $z_1cdots z_k -1$ is irreducible

  3. Consider now $f=z_1cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,hnotinmathbb C$.

  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 $1$?






share|cite|improve this answer









$endgroup$



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 $kge 1$ we assume that $z_1cdots z_k -1$ is irreducible

  3. Consider now $f=z_1cdots z_{k+1}-1$ and assume, by contradiction, that $f=gh$ and $g,hnotinmathbb C$.

  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 $1$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 19 at 12:25









quantumquantum

533210




533210












  • $begingroup$
    Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
    $endgroup$
    – user124697
    Jan 19 at 12:34






  • 1




    $begingroup$
    $f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
    $endgroup$
    – quantum
    Jan 19 at 13:05












  • $begingroup$
    Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
    $endgroup$
    – user124697
    Jan 19 at 13:17










  • $begingroup$
    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?
    $endgroup$
    – user124697
    Jan 19 at 14:36








  • 1




    $begingroup$
    If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
    $endgroup$
    – quantum
    Jan 19 at 15:52


















  • $begingroup$
    Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
    $endgroup$
    – user124697
    Jan 19 at 12:34






  • 1




    $begingroup$
    $f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
    $endgroup$
    – quantum
    Jan 19 at 13:05












  • $begingroup$
    Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
    $endgroup$
    – user124697
    Jan 19 at 13:17










  • $begingroup$
    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?
    $endgroup$
    – user124697
    Jan 19 at 14:36








  • 1




    $begingroup$
    If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
    $endgroup$
    – quantum
    Jan 19 at 15:52
















$begingroup$
Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
$endgroup$
– user124697
Jan 19 at 12:34




$begingroup$
Thank you for answer! However, I didn't fully understand your argument, especially number 4. How can you get the equation $g(z_{1},cdots, 1,z_{k+1}) = cdots = g(1,z_{2},cdots, z_{k})=1$?
$endgroup$
– user124697
Jan 19 at 12:34




1




1




$begingroup$
$f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
$endgroup$
– quantum
Jan 19 at 13:05






$begingroup$
$f(z_1,dots,1,z_k)=g(z_1,dots,1,z_k)h(z_1,dots,z_k)$ 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
$endgroup$
– quantum
Jan 19 at 13:05














$begingroup$
Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
$endgroup$
– user124697
Jan 19 at 13:17




$begingroup$
Oh, now I see. Now I am thinking about 5. Give me a little bit time :)
$endgroup$
– user124697
Jan 19 at 13:17












$begingroup$
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?
$endgroup$
– user124697
Jan 19 at 14:36






$begingroup$
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?
$endgroup$
– user124697
Jan 19 at 14:36






1




1




$begingroup$
If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
$endgroup$
– quantum
Jan 19 at 15:52




$begingroup$
If $gne 1$, what can you say about $g-1$? is it reducible? If so, can you write some of its factors?
$endgroup$
– quantum
Jan 19 at 15:52











1












$begingroup$

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 $mathbb{C}[X_{1}, ldots, X_{n}]/I cong R[X_{n}]/langle fX_{n}-1rangle cong R_{f}$. 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 $a_{n} = frac{1}{a_{1}cdots a_{n-1}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
    $endgroup$
    – user124697
    Jan 21 at 2:42
















1












$begingroup$

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 $mathbb{C}[X_{1}, ldots, X_{n}]/I cong R[X_{n}]/langle fX_{n}-1rangle cong R_{f}$. 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 $a_{n} = frac{1}{a_{1}cdots a_{n-1}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
    $endgroup$
    – user124697
    Jan 21 at 2:42














1












1








1





$begingroup$

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 $mathbb{C}[X_{1}, ldots, X_{n}]/I cong R[X_{n}]/langle fX_{n}-1rangle cong R_{f}$. 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 $a_{n} = frac{1}{a_{1}cdots a_{n-1}}$.






share|cite|improve this answer









$endgroup$



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 $mathbb{C}[X_{1}, ldots, X_{n}]/I cong R[X_{n}]/langle fX_{n}-1rangle cong R_{f}$. 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 $a_{n} = frac{1}{a_{1}cdots a_{n-1}}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 20:34









Alex WertheimAlex Wertheim

16.1k22848




16.1k22848












  • $begingroup$
    Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
    $endgroup$
    – user124697
    Jan 21 at 2:42


















  • $begingroup$
    Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
    $endgroup$
    – user124697
    Jan 21 at 2:42
















$begingroup$
Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
$endgroup$
– user124697
Jan 21 at 2:42




$begingroup$
Wow, this argument let me know about the geometric side of the variety. Thank you very much, I appreciated.
$endgroup$
– user124697
Jan 21 at 2:42


















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