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I know that the zero-set of a non-zero polynomial in $\mathbb{C}[x_1,...,x_n]$ can not have interior points, but I'm trying to find a proof that doesn't require a knowledge of complex analysis like holomorphic functions, etc...

So, suppose that $V=\{(x_1,\cdots, x_n)\in \mathbb{C}^n: p(x_1,\cdots,x_n)=0\}$. Now if there is an interior point in $\mathbf{y}=(y_1,\cdots,y_n) \in V$ then there exists $\epsilon>0$ such that $\displaystyle N_\epsilon(\mathbf{y}) \subseteq V$. But we know that in every metric space open balls are convex, so for all $\lambda \in \mathbb{R}$ the point $\mathbf{\bar{y}}=\lambda\cdot\mathbf{y}+(1-\lambda)\cdot \mathbf{y'} \in N_\epsilon(\mathbf{y})$ lies in $V$where $\mathbf{y'} \in N_\epsilon(\mathbf{y})$.

If I plug $\mathbf{\bar{y}}$ into $p(x_1, \cdots, x_n)=0$ I'll get a new non-zero polynomial $p_1(x_1,\cdots,x_n,\lambda)=0$. Now, if I fix $x_1,\cdots,x_n$ and consider $p_1(x_1,\cdots,x_n,\lambda)$ as a polynomial only in $\lambda$ then I see that $f(\lambda)=p_1(x_1,\cdots,x_n,\lambda)$ then because of convexity $f(\lambda)$ must vanish on every $\lambda \in [0,1]$, but from the fundamental theorem of algebra we know that every non-zero polynomial in $\mathbb{C}[\lambda]$ can have a finite number of roots, so, $f(\lambda)=0$. But I don't see how $f(\lambda)=0$ could lead me to a contradiction.

zyx
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user66733
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3 Answers3

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By induction on $n$, I prove that the only closed subvariety $V$ of $\mathbf C^n$ having $0$ as an interior point is $V=\mathbf C^n$. For $n=1$, it is obvious. Now suppose $V \subseteq \mathbf C^n$ and that $0$ is an interior point of $V$. For each complex hyperplane $H$ passing through $0$, $V \cap H$ is a subvariety of $H \cong \mathbf C^{n-1}$. Moreover $0$ is contained in the interior of $V \cap H$ (relative to the topology of $H$). By the induction hypothesis, it follows that $H = H\cap V$, and since this is true for each $H$ and the collection of hyperplanes covers $\mathbf C^n$, it follows that $V=\mathbf C^n$.

Bruno Joyal
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  • OK. This one looks good and simple, thank you. Only there is one thing I don't understand. Why you chose $0$? I'm new to this subject, so please forgive me if I'm asking trivial questions. – user66733 Oct 02 '13 at 01:35
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    @some1.new4u You're welcome! It's enough to prove it for $0$, because if you want to prove it for another point, you can translate your variety by a linear change of coordinates. I just chose $0$ for notational simplicity. – Bruno Joyal Oct 02 '13 at 01:37
  • Thanks. So, in the mathematical language, linear change of coordinates always gives isomorphic varieties. Right? – user66733 Oct 02 '13 at 01:43
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    @some1.new4u Absolutely. Linear maps are given by (degree 1) polynomials, so they are perfectly acceptable morphisms of algebraic varieties. – Bruno Joyal Oct 02 '13 at 01:45
  • Thanks. It helped a lot. Also, I feel others have given correct answers too, but I don't understand them well because they are somehow ambiguous for me. If it's not too much to ask for, would you please let me know where I am wrong with my proof or understanding their proofs? – user66733 Oct 02 '13 at 01:46
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    @some1.new4u You're welcome. I'm sure the others would love to explain their solutions to you. One is never understood better than by herself! Regards, – Bruno Joyal Oct 02 '13 at 01:49
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    It seems to me that this (wonderfully simple) argument works with $\mathbb{C}$ replaced by any field which is complete with respect to any nontrivial norm. – Pete L. Clark Dec 21 '13 at 22:10
  • @Pete Thank you. Yes, I think that you are right. – Bruno Joyal Dec 21 '13 at 22:22
  • Can we skip the induction and just note that the lines through 0 cover $\mathbb{C}^n$? – Emolga Feb 22 '16 at 10:04
  • @Emolga Yes, you could do that as well. It seems simpler! – Bruno Joyal Apr 29 '19 at 20:17
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We will prove the statement by induction.

The base case $n=1$ follows from the Fundamental Theorem of Algebra, hence we have finitely many solutions and thus no interior points.

For the induction step, supposed that the statement holds for some $k$. We will prove it by contradiction. Suppose $f \in \mathbb{C} [x_1, \ldots x_{k+1}]$ is a polynomial, whose zero set has an interior point $P$. Let $f$ evaluate to $0$ in the ball $B_\epsilon(P)$.

Treating $f$ as a polynomial in $x_{k+1}$, let $f(x_1\ldots , x_{k+1}) = F(x_{k+1})$ be a degree $m$ polynomial. Specifically, let

$$P(x_{k+1}) = \alpha_m x_{k+1}^m + \alpha_{m-1} x_{k+1} ^{m-1} + \ldots + \alpha_0, $$

where each of the $\alpha_i$ are polynomials in $k$ variables, i.e. $\alpha_i \in \mathbb{C} [x_1, \ldots , x_{k}]$. We will now show that $\alpha_i$ must be a polynomial in $k$ variables whose zero set has an interior point, which would contradict the induction hypothesis.

Let $\bar{p}$ denote the projection of the point $p \in \mathbb{C}^{k+1}$ to the first $k$ coordinates. Now, for any $p \in B_\epsilon (P)$, $p = (p_1, p_2, \ldots p_k, p_{k+1})$, since it is also interior point, we can find $m+1$ points in $B_\epsilon (P)$, who agree in the first $k$ coordinates with $p$, where $f$ evaluates to $0$. Let $\alpha_i$ evaluated at $(p_1, \ldots p_{n-1})$ be equal to $ A_i$. Then, we have $k+1$ solutions to the degree $k$ polynomial

$$A_k x^k + A_{k-1} x^{k-1} + \ldots A_0 =0. $$

By the Fundamental Theorem of algebra, these must be a zero polynomial, meaning that $A_i = 0 $. Hence, for all points $p$ in the ball, we have $\alpha_i =0$. Thus, we found a degree $k$ polynomial which evaluates to $0$ in the ball $\overline{B_\epsilon (P) }$, and hence has an interior point. This is our desired contradiction.

user66733
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Calvin Lin
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    Maybe I'm missing something but I think I'm having a problem following your argument. Where did you use the fact that there exists an interior point? Would you write what you're saying rigorously step by step? – user66733 Oct 02 '13 at 00:48
  • @some1.new4u: You should try to follow his hint and expand it into a proof instead. He used the fact $P$ is an interior point to find the $k+1$ points. – Bombyx mori Oct 02 '13 at 00:56
  • @user32240: I have already found infinitely many (actually uncountably many) points. But I'm looking for an explicit reason why that is a contradiction. Please read my proof first if you haven't done so. – user66733 Oct 02 '13 at 01:00
  • @some1.new4u: I have read your "proof", but I think you should spend sometime reading other's real proofs. His is a valid proof and I see no reason you got stuck. – Bombyx mori Oct 02 '13 at 01:01
  • @user32240: I'm really trying to understand his proof, and your proof and I didn't say his proof was wrong. But the problem with his proof is that I don't see where he uses 'interior' point hypothesis. – user66733 Oct 02 '13 at 01:05
  • @some1.new4u: To find the $k+1$ points he has to fix $P$ to be the interior point, find a ball and then pick them. This is the crucial step. – Bombyx mori Oct 02 '13 at 01:07
  • @user32240: Does by any chance he mean $p(\mathbf{P})$ instead of $P(x_n)$? Does it make sense to you to talk about $P(x_n)$ when $P$ is a point? And why does it fail if $P$ is not an interior point? I don't see the things you say explicitly in his answer. Would you please tell me how I can continue with my own proof? because apparently they use the same idea. – user66733 Oct 02 '13 at 01:25
  • @some1.new4u:No offense but I think you should practice your mathematical writing. He means $P(x_{n})$ as a function of $x_{n}$ when $x_{1}\cdots,x_{n-1}$ are fixed. Now he has fixed a point $P$ around which there is a ball such that the polynomial vanishes, then he can obvious pick up $k+1$ points from inside of the both with the same $x_{1}\cdot x_{n-1}$ coordinates. The rest are just induction on the degree of the polynomial $n$. I hope now it is all clear to you. – Bombyx mori Oct 02 '13 at 02:55
  • @some1.new4u: A word of caution: I do not think your "proof" is a proof at all, and his proof used the idea of projecting onto the last coordinate, which is considerably more subtle than yours. I do not think writing "apparently they use the same idea" is appropriate at here. – Bombyx mori Oct 02 '13 at 02:57
  • @some1.new4u I've added the rest of the details, which uses elementary methods (i.e. just FTA), without resorting to complex analysis or topology. – Calvin Lin Oct 02 '13 at 03:58
  • @user32240: I think you haven't read my attempt at all. The idea he is using is the same idea I am using in my attempt to a proof. Moreover, if I claimed that was a proof, I wouldn't have asked for help with it. The idea of projecting onto coordinates is used in my proof when I introduce $f(\lambda)$ to apply FTA on it. I introduced a new variable $\lambda$ to find infinitely many roots for the polynomial $P$ but couldn't go further with $f(\lambda)=0$. His proof wasn't really a proof before he edits his post again. If you want to know what a proof should look like, see Marie's post. – user66733 Oct 02 '13 at 06:39
  • @some1.new4u It's my fault for changing tactics in my edit, because I felt that it provided a clearer understanding of what is going on. The original induction approach that I was thinking of, is that if $f$ is a degree $k_i$ polynomial in each of the variables $x_i$, then if we can find sets $|C_i| = k_i + 1$ such that on each $c = { c_1, c_2, \ldots, c_n| c_i \in C_i}$ (similar to a lattice set of many many points), we have $f(c) = 0$, then $f = 0 $. It's easier to just treat this set as interior points, but clearly will work for the finitely many version too (which is slightly stronger) – Calvin Lin Oct 02 '13 at 13:13
  • @some1.new4u: No, I clearly disagree. Your very words are " Would you please tell me how I can continue with my own proof? because apparently they use the same idea. ". But as you said, yours is not a proof. I have read your post several times, but I do not think you used the projection idea - his proof is to project onto the $x_{n}$ coordinate, not the $\lambda$ coordinate. I wish you be more patient in reading other's work before raising objections like this. – Bombyx mori Oct 02 '13 at 17:05
  • @some1.new4u: I also believe the phrase "His proof wasn't really a proof before he edits his post again." is a bit too strong. His proof before he edit is a bit condensed, but I think he meant you can read what he means and expand it on your own. If you cannot or do not understand that is fine, however asserting "wasn't really a proof" was really too far. – Bombyx mori Oct 02 '13 at 17:19
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One word: differentiate (the defining equations).

It's enough to prove the statement for varieties defined by one equation, because having no interior is a property inherited by subsets.

If the zero set contains an interior point, all derivatives of the equation defining the variety are $0$ at that point (and thus in the a neighborhood of that point). But then the variety defined by any derivative of the equation has the same interior point. By a sequence of differentiations, the highest degree of any term in the defining equation can be reduced to $1$, and the question is reduced to whether a hyperplane has interior points.

If you prefer, differentiate again to reduce to a variety of degree $0$ with no points, therefore no interior points.

zyx
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