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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.

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    $\begingroup$ I don't see why this question was downvoted. It was asked clearly, and the OP showed a good effort. $\endgroup$ Nov 12, 2013 at 0:18

3 Answers 3

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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 $V=\mathbf C^n$.

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  • $\begingroup$ 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. $\endgroup$
    – user66733
    Oct 2, 2013 at 1:35
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    $\begingroup$ @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. $\endgroup$ Oct 2, 2013 at 1:37
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    $\begingroup$ @some1.new4u Absolutely. Linear maps are given by (degree 1) polynomials, so they are perfectly acceptable morphisms of algebraic varieties. $\endgroup$ Oct 2, 2013 at 1:45
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    $\begingroup$ @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, $\endgroup$ Oct 2, 2013 at 1:49
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    $\begingroup$ 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. $\endgroup$ Dec 21, 2013 at 22:10
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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.

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    $\begingroup$ 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? $\endgroup$
    – user66733
    Oct 2, 2013 at 0:48
  • $\begingroup$ @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. $\endgroup$ Oct 2, 2013 at 0:56
  • $\begingroup$ @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. $\endgroup$
    – user66733
    Oct 2, 2013 at 1:00
  • $\begingroup$ @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. $\endgroup$ Oct 2, 2013 at 1:01
  • $\begingroup$ @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. $\endgroup$
    – user66733
    Oct 2, 2013 at 1:05
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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.

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