Choose finite generating sets $I=(g_1,\ldots,g_s)$ and $J=(f_1,\ldots,f_t)$, using the <Hilbert basis theorem>. The assumed inclusion says that each $f_j$ vanishes on $V_{\mathbb C}(I)$. By the , for every $j$ there is an exponent $N_j$ such that
$$
f_j^{N_j}\in I\mathbb C[T_1,\ldots,T_n].
$$

The coefficients in an expression $f_j^{N_j}=\sum_iq_{ij}g_i$ solve a finite <linear system>[system of linear equations] with rational coefficients. Since it has a complex solution, <Gaussian elimination> gives a rational solution. Clearing the finitely many denominators produces a nonzero integer $D$ such that
$$
D f_j^{N_j}\in I
$$
for every $j$.

For any prime $p\nmid D$, reduce these identities modulo $p$. At a common zero of $\pi_p(I)$ in the <algebraic closure> $\overline{\mathbb F}_p$, they give $\pi_p(f_j)^{N_j}=0$, hence $\pi_p(f_j)=0$, for all $j$. Therefore
$$
V_{\overline{\mathbb F}_p}(\pi_p(I))
\subseteq
V_{\overline{\mathbb F}_p}(\pi_p(J))
$$
for every prime except the finitely many divisors of $D$. This is the <spreading out of an affine zero-set inclusion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 3 / b / Solution

Let $I,J\subseteq\mathbb Z[x_1,\ldots,x_n]$. If $V_{\mathbb C}(I)\subseteq V_{\mathbb C}(J)$, then
$$
V_{\overline{\mathbb F}_p}(I\bmod p)\subseteq V_{\overline{\mathbb F}_p}(J\bmod p)
$$
for all but finitely many primes $p$. The Nullstellensatz puts a power of each generator of $J$ in $I\mathbb Q[x_1,\ldots,x_n]$; clearing denominators leaves only finitely many exceptional characteristics.


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