Vanishing Ideals of $\{(t,t)\mid t\in\mathbb{R}\backslash (1,1)\}$ and $\mathbb{Z}^n \subset \mathbb{C}^n$

Question

So for $A :=\{(t,t)\mid t\in\mathbb{R} \backslash (1,1)\}$ in $\mathbb{A}^2(\mathbb{R})$ it seems quite obvious that $I(A)$ should simply be the ideal generated by $x-y$ in $\mathbb{R}[X,Y]$. I tried proving this by sending $f \in I(A)$ to $\mathbb{R}[X]$ with $f(x,y)$ being sent to $g(t):=f(t,t)$ and using the fact that $g$ has infinitely many zeroes. But I'm not sure how to formalize the idea.

As for $B :=\mathbb{Z}^n$ in $\mathbb{A}^n(\mathbb{C})$, I have no clue how one would go about determining $I(B)$. The only fact which seems relevant is that polynomials are holomorphic and therefore analytic, but this doesn't seem to be relevant.

Extra question: what is $V(I(A))$ and $V(I(B))$ where $V$ is the variety of the polynomials.

Thanks in advanced.

Answer 1

For $I(A)$, let $f(x,y)\in I(A)$. Because $$g(t):=f(t,t)\in\mathbb{R}[t]$$ has infinitely many zeros, $g(t)$ is identically zero. That means $f(t,t)=g(t)$ is identically zero. Thus, by treating $f(x,y)$ as an element of the polynomial ring $\big(\mathbb{R}(y)\big)[x]$ in variable $x$ over the field $\mathbb{R}(y)$, we can then see that $f(x,y)$ is divisible by $x-y$. Because $x-y$ is monic in $x$, $$f(x,y)=(x-y)\,q(x,y)$$ for some $q(x,y)\in\big(\mathbb{R}[y]\big)[x]=\mathbb{R}[x,y]$. Therefore, $$I(A)=\langle x-y\rangle\,.$$ Thus, $$V\big(I(A)\big)=\big\{(t,t)\,\big|\,t\in\mathbb{R}\big\}\,.$$

As for $I(B)$, we use the theorem below. Since each $f(z_1,z_2,\ldots,z_n)\in I(B)$ vanishes on $B=\mathbb{Z}^n$. with $\mathbb{Z}$ being an infinite subset of $\mathbb{C}$, we get that $f(z_1,z_2,\ldots,z_n)=0$ identically. Thus, $$I(B)=\{0\}\,,$$ and so $$V\big(I(B)\big)=\mathbb{C}^n\,.$$

Theorem. Let $\mathbb{K}$ be a field and $n$ a positive integer. Suppose that a polynomial $f(z_1,z_2,\ldots,z_n)\in\mathbb{K}[z_1,z_2,\ldots,z_n]$ vanishes on $S_1\times S_2\times\ldots\times S_n$, where $S_j$ is an infinite subset of $\mathbb{K}$. Then, $f(z_1,z_2,\ldots,z_n)$ is the zero polynomial.

For a proof, we work by induction on $n$. If $n=1$, the claim is trivial. Suppose that $n>1$. Write $$f(z_1,z_2,\ldots,z_n)=\sum_{r=0}^d\,f_r(z_1,z_2,\ldots,z_{n-1})\,z_n^r\,,$$ where $d$ is the degree of $z_n$ in $f(z_1,z_2,\ldots,z_n)$, and $$f_r(z_1,z_2,\ldots,z_{n-1})\in \mathbb{K}[z_1,z_2,\ldots,z_{n-1}]$$ for $r=0,1,2,\ldots,d$. Fix an arbitrary tuple $$(s_1,s_2,\ldots,s_{n-1})\in S_1\times S_2\times\ldots\times S_{n-1}\,.$$ Then, $$f(s_1,s_2,\ldots,s_{n-1},z_n)=\sum_{r=0}^d\,f_r(s_1,s_2,\ldots,s_{n-1})\,z_n^r\,,$$ as a polynomial in $z_n$ has infinitely many roots (as it vanishes on $S_n$). Thus, the polynomial $f(s_1,s_2,\ldots,s_{n-1},z_n)$ is identically zero as a polynomial in $z_n$. Ergo, we have $$f_r(s_1,s_2,\ldots,s_{n-1})=0$$ for all $r=0,1,2,\ldots,d$. By induction hypothesis, each $f_r(z_1,z_2,\ldots,z_{n-1})$ is identically zero. Thus, $f(z_1,z_2,\ldots,z_n)$ is also the zero polynomial.

Remark. If $d_j$ is the degree of $z_j$ in $f(z_1,z_2,\ldots,z_n)$ for each $j=1,2,\ldots,n$, then it suffices to only require that $|S_j|>d_j$ for $j=1,2,\ldots,n$. The assertion of the theorem above is still true.

Answer 2

For B:

We want to prove that $\Bbb Z^n$ is Zariski-dense in $\Bbb C^n$, i.e. that the only polynomial that vanishes on $\Bbb Z^n$ is the zero polynomial.

Let $\Lambda\simeq{\Bbb Z}^n$ and $V=\Lambda\otimes{\Bbb C}\simeq{\Bbb C}^n$.

If we denote $\Lambda^\ast={\rm Hom}(\Lambda,{\Bbb Z})$ and $V^*={\rm Hom}(V,{\Bbb C})$ the dual spaces there are isomorphisms $$ \Lambda^\ast\otimes{\Bbb C}\simeq{\rm Hom}(\Lambda,{\Bbb C})\simeq V^\ast $$ because a linear form on a vector space is completely determined by its value on a basis. Thus, if $\rm Sym^\bullet$ denotes the symmetric algebra the restriction $$ {\rm Sym}^\bullet(V^\ast)\longrightarrow{\rm Sym}^\bullet({\rm Hom}(\Lambda,{\Bbb C})) $$ defined by the inclusion $\Lambda\subset V$ is actually an isomorphism. In particular a symmetric form on $V$ restricts to $0$ on $\Lambda$ if and only if it was null on $V$ in the first place.

This is enough because ${\rm Sym}^\bullet(V^\ast)\simeq{\Bbb C}[z_1,...,z_n]$.