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In general, if $K$ is a field, it could be that exists $f(x)\in K[x]$ such that $f(a)=0$ for all $a\in K$; for example, set $K:=\mathbb Z/(2)$ and $f(x):=x^2+x$.

Now, if $f(x)$ vanishes on all $\operatorname{Spec} K[x]$, it means that $f(x)\in \mathfrak p$ for all $\mathfrak p\subset K[x]$, so that $f(x)$ is nilpotent. However $1$ is clearly not nilpotent. Is this because the sets $\operatorname{Spec} K[x]\neq K$ in general, and the fact that $f$ vanishes on all $K$ only means that $f$ belongs to the (always maximal) ideals of the form $(x+a)$ for $a\in K$? That would make sense to me, in fact in the example above $x^2+x$ belongs to both $(x)$ and $(x+1)$.

So is there any characterization of the fields $K$ for which, if $f\in K[x_1,\dots ,x_n]$ vanishes on all $K^n$, it vanishes on $\operatorname {Max Spec}K[x_1,\dots ,x_n]$? For example if $K$ is algebraically closed and $n=1$, then as sets $\operatorname{Max Spec}K[x_1]=K$. However this already isn't true anymore for $n\gt 1$ or if $K$ is not algebraically closed. For example, are there any polynomials in $\mathbb R[x]$ belonging to all the ideals of the form $(x-a)$ but not in $x^2+1$? What is the geometric interpretation of this fact, if there is any?

Dr. Scotti
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    I'm confused by what you mean by $\operatorname{Spec} K[x] = K$. This does not make sense as $\operatorname{Spec} K[x]$ is a scheme (and hence a locally ringed space) and $K$ is a commutative ring with identity, so they cannot be equal (one is a ring and the other is a topological space equipped with a structure sheaf). Do you mean that the closed points of $\operatorname{Spec} K[x]$ are in bijection (or not) with the elements of $K$? – Geoff Jan 11 '22 at 15:11
  • My bad sorry. For $\operatorname {Spec} K[x]$ I mean just the set of prime ideals equipped with the Zariski topology, and when I said that it is equal to $K$ I meant that its underlying set is in bijection with $K$ (that is not correct though, it is $K$ plus the point corresponding to $(0)$). – Dr. Scotti Jan 11 '22 at 15:20

2 Answers2

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What you ask happens exactly when $K$ is infinite. If $K=\Bbb F_q$ is finite, just take $\prod_{a\in K} (x_1-a)$ which vanishes on every element of $\Bbb F_q^n$ but not on $(b,0,\cdots,0)$ where $b\in\Bbb F_{q^2}\setminus\Bbb F_q$. If $K$ is infinite, you may prove by induction that the only polynomial which is zero on all of $K^n$ is zero: write $p(x_0,\cdots,x_n)$ as a polynomial in $x_n$ with coefficients in $K[x_0,\cdots,x_n]$, then see that if there's any point $(a_1,\cdots,a_{n-1})$ so that the coefficients don't all vanish we have $p$ has only finitely many roots of the form $(a_1,\cdots,a_{n-1},b)$ as $b$ varies. So all the coefficients must vanish, and eventually you're reduced to the case of polynomials in $K[x]$ vanishing at all elements of $K$: but every polynomial in one variable has finitely many roots over a field.

The geometric interpretation is that the points $K^n\subset \operatorname{Spec} K[x_1,\cdots,x_n]$ are dense iff $K$ is infinite.

Hank Scorpio
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  • I must be dense, but I don't understand the two proofs (the general picture is clear though). For instance, why do we take $\mathbb F_{q^2}$? Shouldn't we search for an irreducible polynomial $p\in\mathbb F_q[x]$ not divisible by that polynomial $f$ obtained with the productory? In fact (even if I found your geometric obsevation interesting) I meant to ask the interpretation of when an $f\in K[x]$ vanishes on all $K^n$ but not on another $p\in \operatorname {Max Spec}K[x]$.Even in the second proof, I'm not understanding your argument (also the index confuse me..) Thanks for your patience – Dr. Scotti Jan 11 '22 at 21:13
  • You can take any algebraic extension, I just did $\Bbb F_{q^2}$ because it's the smallest one. Your next logic is a little off - $(x_1-a)$ is a polynomial not divisible by the product; what you really want is a polynomial coprime to the product, and you can take the minimal polynomial of $b$ which is coprime to the product because $b-a_i\neq 0$ for all $i$. The second proof is standard: for any fixed $(a_1,\ldots,a_{n-1})$, the polynomial $p(a_1,\cdots,a_{n-1},x_n)$ is a polynomial in a single variable which must have infinitely many roots, so it must be 0. Therefore the coefficients vanish – Hank Scorpio Jan 11 '22 at 22:59
  • for all choices of $a_1,\ldots,a_{n-1}$ and so by applying this again to each of the coefficients of $p$ viewed as a polynomial in $x_n$, we eventually find that all of those are zero and so $p=0$. I'm sure this is in lots of textbooks; if you're still having trouble I recommend going and looking at one. Or here on MSE. – Hank Scorpio Jan 11 '22 at 23:01
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For $K$ a field $K[X_1,…,X_n]$ is an integral domain, in particular there are no nilpotent elements besides $0$. Your argument shows that if $f(x_1,…,x_n)=0\in K$ for all choices of $x_1,…,x_n \in K$, then we have $$f \in \bigcap \limits_{{x}\in K^n} (X_1-x_1,…,X_n-x_n)$$ If $K$ is algebraically closed this is by Hilbert’s Nullstellensatz equivalent to saying that $$f \in \bigcap \limits_{\mathfrak{m}\in \operatorname{mSpec}(K[X])} \mathfrak{m} = \operatorname{jrad}(K[X])$$ which is the jacobson radical.

I don’t see why you should be able to deduce $f$ being in all prime ideals. I think in the case of $K[X]$ with $K$ algebraically closed this only works since all prime ideals are already maximal.

Jonas Linssen
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