Geometric meaning of $\operatorname{Spec}(k[x_1,\ldots, x_n])$ 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.
 A: 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 $x_1, ... , x_n$.
A: The association is $(a_1, \dots, a_n) \to (x_1 - a_1, \dots, x_n - a_n)$. The subspace topology of maximal ideals then coincides with the Zariski topology on the points.
For your second question, let
$$A = k[x,y]/(x^3 - y^2 +xy) \; .$$
By setting $t = y/x$ we get $t(t-1) = \frac{y^2 - yx}{x^2} = x$, and $t^2(t-1) = \frac{y^3 - y^2 x}{x^3} = y \frac{x^3}{x^3} = y$
$$\overline{x} = t(t-1), \overline{y} = t^2 (t - 1) \; ,$$
so $A \cong k[t(t-1), t^2 (t-1)]$.
Now, recall that the Spec of $R/I$ is $V(I)$, the set of prime ideals containing $I$. In our case, the maximal ideals containing $I  = (x^3 - y^2 + xy) = (f(x, y))$ are precisely of the form $(x - a, y - b)$ as you remarked (corresponding to points), but this tells us that such points $(a, b)$ satisfy this equation of the contained ideal, since $f(x, y) = p(x, y)(x - a) + q(x, y)(y - b)$, thus we have an association with the curve.
By the way, this is just my preference but I read a little of Matsumura and found it hard for a first time in commutative algebra, I prefer Atiyah-Macdonald much more.
