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Let $\mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$.

Is it possible to find a norm on $\mathbb{C}[x_1,\ldots,x_n]$? (If not, why?).

Where a norm $\rho$ is a function $\rho: \mathbb{C}[x_1,\ldots,x_n] \to \mathbb{R}$, having the following properties: $\forall c \in \mathbb{C}, u,v \in \mathbb{C}[x_1,\ldots,x_n]$,

(1) $\rho(v) \geq 0$. $\rho(v)=0$ if and only if $v=0$

(2) $\rho(cv)= |c|\rho(v)$

(3) $\rho(u+v) \leq \rho(u)+\rho(v)$

Any comments are welcome!

I have a problem with the second property, which ruins some ideas I had in mind.

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  • $\begingroup$ There certainly is a norm on this ring given your conditions. I don't know if there's a nice, clean one, but one does exist. $\endgroup$ – Matt Samuel Apr 25 '17 at 2:30
  • $\begingroup$ @MattSamuel, thanks!! But how do you know that? Is there an existence theorem that I am missing? $\endgroup$ – user237522 Apr 25 '17 at 2:32
  • $\begingroup$ Wouldn't​ the norm of the coefficient treated as a vector work? $\endgroup$ – Jean-François Gagnon Apr 25 '17 at 2:37
  • $\begingroup$ @Jean Yeah, you could take the square root of the sum of squared absolute values of the coefficients. That should work. $\endgroup$ – Matt Samuel Apr 25 '17 at 2:40
  • $\begingroup$ But what about $n \geq 2$? $\endgroup$ – user237522 Apr 25 '17 at 2:41
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The polynomial structure is a red herring; you only care about its vector space structure, and $\mathbb{C}[x_1, \ldots, x_n]$ is a complex vector space of countable dimension.

Another description of this space is that it is the vector space of all complex sequences that have only finitely many nonzero terms.

As an example, any of the $L^p$ norms would work: for real $p \geq 1$

$$ \| \vec{a} \|_p = \left( \sum_{n} | a_n |^p \right)^{1/p} $$

defines a norm, as does,

$$ \| \vec{a} \|_{\infty} = \max_n\left( |a_n| \right) $$

Basically the same description works to define these norms on polynomials directly, rather than the alternate representation in terms of sequences: just take the sums over the coefficients of the polynomial rather than terms of a sequence.

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  • $\begingroup$ Thanks! Please, how one shows that $\mathbb{C}[x_1,\ldots,x_n]$ is not complete in each of those norms? $\endgroup$ – user237522 Apr 25 '17 at 3:15
  • $\begingroup$ @user237522 with this norm $\mathbb{C}[x_1, \ldots, x_n]$ is isomorphic to $\mathbb{C}^n$ with its $L^p$ norm. Thus this normed vector space is complete. $\endgroup$ – Robert Wolfe Apr 25 '17 at 4:04
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    $\begingroup$ @Bryan No, it isn't. $\mathbb{C}[x_1, \ldots, x_n]$ is infinite-dimensional since it contains all polynomials in $n$ variables. You are confusing the number of variables with dimension. $\endgroup$ – Conifold Apr 25 '17 at 4:21
  • $\begingroup$ @user237522 You can use the Baire category theorem to show that any infinite Banach space must have uncountably infinite dimension. Otherwise, let $\{a_1, \ldots \}$ be a basis. Let $V_n$ be the span of the first n basis vectors. Then each $V_n^c$ is dense and open. If the space was complete, the intersection all of the $V_n^c$ would be dense by the Baire category theorem. However, you can easily see that this intersection is empty, since every vector (which is a finite linear combination of the $a_i$, we are talking basis as a vector space) must lie in some $V_n$. $\endgroup$ – Lorenzo Najt Apr 25 '17 at 5:14
  • $\begingroup$ @Conifold Are we talking about $\mathbb{C}[x_1, \ldots, x_n]$ for a fixed $n$ or $\bigcup_{n=1}^\infty \mathbb{C}[x_1, \ldots, x_n]$? The OP doesn't make any remark claiming that a polynomial of arbitrarily large strings of variables may occur. $\endgroup$ – Robert Wolfe Apr 25 '17 at 12:56

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