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While reading the Gortz's Wedhorn's Algebraic Geometry, proof of Theorem 13.89, I encountered next situation : Let $C\subsetneq \mathbb{P}^{m}_k$ - the notation $\mathbb{P}^m_k$ means the projective space regarded as a scheme - be a proper closed subscheme of a projective space over an 'infinite' field $k$. Then does there exists a $k$-valued point $q\in \mathbb{P}^{m}(k)$ - the notation $\mathbb{P}^{m}(k)$ means the projective space regarded as a classical variety - which is not contained in $C$ ( because there exists no nonzero homogeneous polynomial that vanishes on all points of $\mathbb{P}^{m}(k)$ ? )? Can we try to use the nonexistence of such homogeneous polynomial? If all $k$-valued points are contained in $C$, then such homogeneous polynomial exists? If our question is not true in general, how about the case that $C:=\operatorname{Im}(f)$, where $f: X \to \mathbb{P}^{m}_k$ is an nonsurjective closed immersion ? Since $C$ is proper subset, we can choose an element in $\mathbb{P}^{m}_k - C$ but I don't know how to choose it as a k-valued point.

EDIT : Consider next situation. Let $\mathbb{P}^{m}(k)$ be a projective space over infinite field $k$, which is considered as the classical variety. Let $C \subsetneq \mathbb{P}^{m}(k)$ be a proper ( Zariski ) closed subset. Then $C$ is of the form $V_{+}(f_1, \dots , f_r)$ for some homogeneous polynomials $f_1, \dots ,f_r \in k[X_0, \dots , X_m]$.

Interlude question : In this setting, the $k$-valued point of $\mathbb{P}^{m}(k)$ means just a point whose homogeneous coordinates are elements of the field $k$ ?

Assume this question is true. Since $C= V_{+}(f_1, \dots, f_r)$ is a proper subset, we assume that all $f_i$ are nonzero. By the bold statement in our question - there exists no nonzero homogeneous polynomial that vanishes on all points of $\mathbb{P}^{m}(k)$ since $k$ is infinite - , for each $1 \le i \le r$, there exists $q_i \in \mathbb{P}^{m}(k)$ such that $f_i(q_i) \neq 0$. In particular, $f_1(q_1) \neq 0$ so that $q_1 \notin C$. So if the interlude question is true, then $q_1$ is $k$-valued point ( True? I do not completely convince for this part ). So I think we find such $q \in \mathbb{P}^{m}(k)$ as in the above bold statement.

But let's consider different direction. If the interlude question is true, then since $C$ is proper subset any $q\in \mathbb{P}^{m}(k)-C$ satisfies such condition in the bold statement? But why Gortz, Wedhorn did not argue directly as this? In this second argument, it seems that the condtion that $k$ is infinte is redundant but I think we should use the infinity of $k$. Where the infiniteness of $k$ is used exactly? Is there a problem in the above interlude question? Is there a point that I missing something? I'm a little bit confused. Can anyone please clarify this issue a bit more?

Plantation
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    Your question is local and it basically asks if a finitely generated $k$-algebra $A \neq 0$ has any maximal ideal whose quotient is $k$. But consider for example $\mathbb{Q}[X] / (X^2 + 1)$. – L Y Jun 23 '25 at 05:25
  • @LY : O.K. Just consider $\varnothing \neq U := \mathbb{P}^{n}_k -C$. By considering affine open covers, there exists $A\neq0$ which is a finitely generated $k$-algebra such that $\operatorname{Spec}A \subseteq U$. So if there exists a maximal ideal $x =\mathfrak{m}_x \subseteq A$ such that $\kappa(x)=A/\mathfrak{m}_x = k$, then $k\hookrightarrow \kappa(x)$ is an isomorphism so that $x$ is a $k$-valued ( or rational ) point. And your comment is true? Your example is a counter example?.. – Plantation Jun 23 '25 at 06:07
  • @LY : And the bold sentence in our question, "because there exists no nonzero homogeneous polynomial that vanishes on all points of $\mathbb{P}^{m}(k)$", is redundant in showing our goal of finding such $k$-valued point ? There is a possibility that we should use other method to find such suitable $k$-valued point, than my previous comment. – Plantation Jun 23 '25 at 06:08
  • @LY : My understanding about your example is, since $\mathbb{Q}[X]/(X^2+1)$ is a field, $(0)+(X^2+1)$ is its only maximal ideal, and its quotient is $\mathbb{Q}[X]/(X^2+1)$, same (?) , which is different to $\mathbb{Q}$. So the claim you exhibited may false. I think we'll have to try another way. Can we breakthrough this difficulty? – Plantation Jun 23 '25 at 07:23
  • Are you asking why for any polynomial $f$ there is a $k$-point $p$ with $f(p)\neq 0$? This has been answered, for example, here: https://math.stackexchange.com/q/2789699/1213739 – Leobeth Jun 23 '25 at 07:54
  • @Leobeth : I think it is different from my question. The statement you exhibit seems work from the infinity of $k$. Question that I ask is, "if all $k$-valued points of the projective space $\mathbb{P}^{m}(k)$ are in $C$, then does there exists a nonzero homogeneous polynomial that vanishes on all points of $\mathbb{P}^{m}(k)$, which leads to contradiction." Can you help me? – Plantation Jun 23 '25 at 08:01
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    Yes, pick any non-zero homogeneous polynomial from the defining ideal $I(C) \subseteq k[x_0,\dots,x_m]$. – Leobeth Jun 23 '25 at 08:06
  • O.K. Pick homogeneous $0 \neq p\in I(C)$. How can we prove that such $p$ vanishes on all points of $\mathbb{P}^{m}(k)$, from that all $k$-valued points are in $C$? If you are familiar to scheme language, I accept the 'notion of $k$-valued point' as a point $x\in \mathbb{P}^{m}(k)$ such that the natural map $k \hookrightarrow \kappa(x)$ is an isomorphism. Can we make proof more formally? – Plantation Jun 23 '25 at 08:26
  • And P.s. Again if you are familiar to scheme language, I regard $\mathbb{P}^{m}(k)$ as the projective spectrum $\operatorname{Proj}k[T_0, \dots, T_m]$. In this setting, what is the defining ideal $I(C)$? I understand your intention roughly and I can't translate your strategy to the scheme language. If you have time, can you help? – Plantation Jun 23 '25 at 08:34
  • Even if we are working in the language of classical varieties, in this case, I don't know what $k$-valued point exactly means.. – Plantation Jun 23 '25 at 08:38
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    If $\mathbb P^m = \operatorname{Proj} k[x_0,\dots,x_m]$, or more generally any affine or projective variety $X$ over $k$, then $X(k)$, the maps $\operatorname{Spec} k \to X$, coincides exactly with (classical) points with coordinates in $k$. See Example (5.3) in GW. – Leobeth Jun 23 '25 at 09:05
  • @Leobeth : O.K. Let's back to the bold statement in my question. As Example (5.3) in G.W. , if $k$-valued points of $\mathbb{P}^m_k$ are in bijection with $\mathbb{P}^{m}(k)$ ( classical points with coordinates in $k$ ), then since $C \subsetneq \mathbb{P}^{m}(k)$, any element in $\mathbb{P}^{m}(k) -C$ is a $k$-valued point which is not contained in $C$ and we are done (?). Why the G.W. mentioned that "because there exists no nonzero homogeneous polynomial that vanishes on all points of $\mathbb{P}^{m}(k)$" ? Where such sentence is used? Is there a hole in my argument? Did I miss something? – Plantation Jun 23 '25 at 10:01
  • @Leobeth : I have organized my questions and posted them as in EDIT. Perhaps, can you see ? – Plantation Jun 24 '25 at 03:24

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