The A-valued points on a scheme

Question

Let $A$ be a ring and $X=Spec(\mathbb{Z}[x_1,...,x_n] / (f_1,...,f_r))$ an affine scheme where $f_i \in \mathbb{Z}[x_1,...,x_n]$. The $A$-valued points of $X$ are defined to be morphisms $Spec(A)\rightarrow X$. Show that the $A$ valued points of $X$ are the solutions to $f_1(x_1,...,x_n)=\cdots=f_r(x_1,...,x_n)=0$ in the ring $A$.

Answer 1

If $A$ is a ring, also known as a $\mathbf{Z}$-algebra, then ring maps from $\mathbf{Z}[X_1,\ldots,X_n]$ to $A$ are the same as $n$-tuples of elements of $A$. This is the universal property of the polynomial ring:

$\mathrm{Hom}_{\mathbf{Ring}}(\mathbf{Z}[X_1,\ldots,X_n],A)=A^n$

via $\varphi\mapsto(\varphi(X_1),\ldots,\varphi(X_n))$. A ring map $\varphi:\mathbf{Z}[X_1,\ldots,X_n]\rightarrow A$ factors through $\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m)$ if and only if each $f_i$ is sent to zero. By definition, $\varphi(f_i)=f_i(a_1,\ldots,a_n)$, where $a_i=\varphi(X_i)$. Conversely, any ring map $\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m)\rightarrow A$ gives rise to a homomorphism $\mathbf{Z}[X_1,\ldots,X_n]\rightarrow A$ which sends each $f_i$ to zero. Thus

$\mathrm{Hom}_{\mathbf{Ring}}(\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m),A)\hookrightarrow \mathrm{Hom}_{\mathbf{Ring}}(\mathbf{Z}[X_1,\ldots,X_n],A)=A^n$

identifies the source (as a set) with the set $n$-tuples $(a_1,\ldots,a_n)\in A^n$ such that $f_i(a_1,\ldots,a_n)=0$ for $1\leq i\leq n$.

It remains to translate this to a statement about the $A$-valued points of $\mathrm{Spec}(\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m))$. This is done via the $\mathrm{Spec}$ functor, which gives an equivalence of categories between rings and affine schemes. So ring maps from $\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m)\rightarrow A$ are the same as morphisms $\mathrm{Spec}(A)\rightarrow\mathrm{Spec}(\mathbf{Z}[X_1,\ldots,X_n]/(f_1,\ldots,f_m))$.