I am a beginning learner of commutative algebra, using the book commutative algebra by Matsumura. In the book he often refers $\operatorname{Spec}(k[x_1,\ldots, x_m])$ to an affine plane where $k$ is a field. But I do not understand how this identification works. Is there a canonical isomorphism (or homeomorphism) between $\operatorname{Spec}(k[x_1,\ldots,x_n])$ and $\mathbb A^n(k)$?
I understand that when $k$ is algebraically closed, then by Hilbert Nullstellensatz we can identify the maximal ideals with points in $k^n$ but not sure if it is related to this identification of prime ideals and $\mathbb A^n(k)$.
Moreover, the book has an example on $k[x]$: If we put $x_1=x(x-1)$ and $x_2=x^2(x-2)$ then $\operatorname{Spec}(k[x_1, x_2])$ is the affine curve $x_1^3-x_2^2+x_1x_2=0.$ I kinda see this is the relation that $x_1$ and $x_2$ satisfies but still do not see how the prime ideals can be associated with the curve formally. I guess this is a elementary and standard picture in one's mind but I do not see a mapping which makes this picture formal as a beginner. Please correct me or provide me with any sort of insights.
Best Answer
I like the functor of points perspective. You want to think of affine space as $k^n$, or more generally $R^n$ for a ring $R$. The "universal affine space" $\operatorname{Spec} k[x_1, ... , x_n]$, as a scheme over $k$, has the property that for any $k$-algebra $R$, there is a natural bijection of sets
$$\operatorname{Hom}_{\textrm{$k$-schemes}}(\operatorname{Spec} R, \operatorname{Spec} k[x_1, ... , x_n]) = R^n. $$
This is because the set on the left is naturally in bijection with
$$\operatorname{Hom}_{\textrm{$k$-alg}}(k[x_1, ... , x_n], R)$$
and a $k$-algebra homomorphism is completely determined by where it sends $x_1, ... , x_n$.