Geometric view of isomorphism between sets My studying of Commutative Algebra lead me to the following question (though not a strictly Commutative Algebra question) :
What is a (or the) geometric intuition of the isomorphism
$$\mathbb{R}[x,y]/(x^2-y,x-x^3+xy)\simeq\mathbb{R}$$
So $\mathbb{R}[x,y]$ is the ring of polynomials with variables $x,y$ and $(x^2-y,x-x^3+xy)$ is an ideal of $\mathbb{R}[x,y]$.
I would like to see an awnser for the spesific question for I have searched for 'geometric approaches' of isomorphisms in classics like the Atiyah MacDonald 'Introduction to Commutative Algebra' and in Algebraic Geometry material and I cannot figure out the idea. Really tough, I would like to know your mind considering geometric approaches of isomorphisms 'in general' (perhaps by giving an example that you have in mind, like the one above and of course, the more examples, the better). Also, any sources for further studying or links to previously related posts are welcome.
 A: The short answer: the curves $x^2 - y = 0$ and $x-x^3+xy = 0$ intersect in $1$ point, namely $(0,0)$. Correspondingly, the ring $\mathbb{R}[x,y]/(x^2 - y, x-x^3+xy)$ is isomorphic to $\mathbb{R}$ via the map sending $x \mapsto 0$ and $y \mapsto 0$. Note that

*

*$\mathbb{R}$ has Krull dimension $0$, which reflects that the intersection (a point) is a $0$-dimensional variety; and

*$\mathbb{R}$ is a $1$-dimensional $\mathbb{R}$-vector space, corresponding to the number of intersection points of the curves. (In general, the dimension is the number of points counted with multiplicity.)

The longer answer: Let $f_1 = y - x^2$ and $f_2 = x-x^3+xy = x(y - (x^2 - 1))$. To simplify things in the beginning, let’s work over $\mathbb{C}$, which is algebraically closed. The vanishing locus of $\{f_1, f_2\}$ (or equivalently of the ideal $(f_1, f_2)$) is the set $\newcommand{\V}{\mathbb{V}} \newcommand{\C}{\mathbb{C}} \mathbb{V}(f_1,f_2)$ of points $(x_0, y_0) \in \C^2$ such that
$$
f_1(x_0,y_0) = f_2(x_0,y_0) = 0 \, .
$$
Given a point $(x_0, y_0) \in \mathbb{C}^2$, then $(x_0, y_0) \in \V(f_1,f_2) \iff (f_1, f_2) \subseteq (x - x_0, y - y_0)$; see here for a proof. By the correspondence theorem for ideals in a quotient, this means that $(x - x_0, y - y_0)/(f_1, f_2)$ is an ideal, and in fact a maximal ideal of the coordinate ring $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$. Moreover, all such maximal ideals of $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$ are of this form by the Nullstellensatz. Thus we get an algebro-geometric dictionary in which the points of the algebraic set given by $f_1 = f_2 = 0$ are in bijection with the maximal ideals of the coordinate ring $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$.
Over a non-algebraically closed field like $\mathbb{R}$, things are a bit more complicated. Maximal ideals of the coordinate ring $\frac{\mathbb{R}[x,y]}{(f_1, f_2)}$ don’t correspond to $\mathbb{R}$-points of $\V(f_1,f_2)$, but instead to Galois orbits of $\mathbb{C}$-points. For instance, since there are no real solutions to the equation $x^2 + 1 = 0$, the maximal ideal $(x^2 + 1)$ of $\mathbb{R}[x]$ doesn’t correspond to any real point. Instead it corresponds to the Galois orbit $\{i, -i\}$ of complex solutions to the equation, since $x^2 + 1 = (x-i)(x+i)$.
Fortunately in your example things are simple. The vanishing locus of $f_1 = x^2 - y$ is the parabola $y = x^2$, and the vanishing locus of $f_2 = x(y - (x^2 - 1))$ is the union of the $y$-axis and the parabola $y = x^2 - 1$. As you can see in the plot below, the intersection of these two algebraic sets is simply the point $(0,0)$.
$\hspace{2.5cm}$
Algebraically, one can see this by noting that $(f_1, f_2) = (x,y)$, since
$$
x = f_2 + x f_1 \qquad \text{and} \qquad y = f_1 + x^2 = f_2 + x(f_2 + x f_1) \, .
$$
Thus
\begin{align*}
\frac{\mathbb{R}[x,y]}{(f_1, f_2)} = \frac{\mathbb{R}[x,y]}{(x,y)} \cong \mathbb{R}
\end{align*}
where the last isomorphism takes $x \mapsto 0, y \mapsto 0$.
For similar connections between algebra and geometry, see this post and this post. For more on the algebro-geometric dictionary and how geometric properties are reflected in the corresponding algebraic structures, see this post.
A: I think you might want to get a better idea of algebraic quotients in general, here's an answer of mine where I try to go in depth for a beginner: Ideal generated by $v\otimes v - \Phi(v)1$
Most introductions to category theory give a good overview of quotients; I'd suggest Emily Riehl's Category Theory in Context.
The core geometric insight is that for any algebraic quotient, the kernel of the quotient homomorphism (the ideal) is reduced to $0$, and any affine space parallel to that ideal is also reduced to a single point (but nonzero).
If you really in need of a visualization, I'd suggest taking a 3D graphing calculator like GeoGebra, and graphing $f(x, y) = x^2 - y^2 + C$, making $C$ vary, and trying to understand how reducing these surfaces to a point (for each $C$) will draw a line/curve. Though this is not at all your example, it should give you an analogy of what's happening; your example is much simpler to understand algebraically than it is to visualize geometrically.
Edit: a previous version of this answer mistakenly claimed that $\mathbb{R}[x,y]/(x^2−y, x−x^3+xy)$ was a curve, when it is, in fact, a point. This has been corrected; and @Viktor Vaughn's answer accepted as correct. The points I make on algebraic quotients still hold, though, so I have left them unchanged.
A: Here's how I think about it. $\mathbb{R}[x, y]$ is the ring of all polynomial functions on $\mathbb{R}^2$ under pointwise addition and multiplication. The "$\mathbb{R}$" on the right-hand side of your isomorphism is not some abstract version of the real field. It's actually all polynomial functions whose domain is the single point $(0, 0) \in \mathbb{R}$.
Thought of in these terms, the restriction from $\mathbb{R}^2$ to $\{(0, 0)\}$ corresponds to the obvious surjective map $\mathbb{R}[x, y] \to \mathbb{R}$ given by $x \mapsto 0, y \mapsto 0$ that just picks off the constant term. This is a ring homomorphism because function arithmetic is defined pointwise, which restriction clearly respects.
As for the ideal $(x^2-y, x-x^3+xy)$, passing to the quotient $\mathbb{R}[x, y]/(x^2-y, x-x^3+xy)$ literally says we want to consider all polynomial functions on $\mathbb{R}^2$ with the additional relations that $x^2-y = 0$ and $x-x^3+xy = 0$ but no others except those implied by these. The claimed isomorphism is a concrete geometric way to think of what this abstract quotient "looks like": if it's true, it says that any two polynomials in $\mathbb{R}[x, y]$ with the same constant term can be shown to be equal if you allow yourself to write $x^2-y=0$ and $x-x^3+xy=0$ in intermediate steps, and that there is no additional "collapsing" beyond this.
To prove it, first off the two relations respect restriction to $\{(0, 0)\}$ since $0^2 - 0 = 0$ and $0-0^3+0\cdot0 = 0$, so restriction continues to give a surjective map $\mathbb{R}[x, y]/(x^2-y, x-x^3+xy) \to \mathbb{R}$. It is natural to ask if this map is injective as well. Geometrically, that's equivalent to asking: having imposed the two abstract relations above and no more, can I have two different polynomial functions which have the same value at $(0, 0)$?
Before answering that question, let's consider what would happen if $x^2-y$ and $x-x^3+xy$ were also both zero at some point $(a, b)$ different from $(0, 0)$. Then we could pick polynomials $p(x, y), q(x, y)$ such that $p(0, 0) = q(0, 0)$ yet $p(a, b) \neq q(a, b)$. Using the restriction to $\{(a, b)\}$ rather than $\{(0, 0)\}$, that would force $p \neq q$ in the quotient even though their constant terms are the same, so the map would not be injective. Hence more common zeros keep the quotient bigger. However, it's easy to see there are no common zeros here beyond $(0, 0)$, so this obstruction doesn't appear in this case.
It's inevitable at this point that we'll have to start playing with algebra. For instance, we get the same obvious surjection $\mathbb{R}[x, y]/(x^2, y^2) \to \mathbb{R}$ by restricting to $\{(0, 0)\}$, but the quotient is larger. (The map which takes the partial derivative with respect to $x$ and then evaluates at $(0, 0)$ is well defined.)
Ok, let's play. We've got $y=x^2$ and $xy=x^3-x$. If we could divide by $x$ the second would be $y=x^2-1$ and then we'd have imposed $0=-1$ which would result in the zero ring and everything would collapse completely. But we can't divide by $x$, roughly since that might mean dividing by $0$. But if we multiply $y=x^2$ by $x$ we can do almost the same thing: $xy = x^3$ and $xy = x^3-x$. Together this says $x=0$, but since $y=x^2$ we get $y=0$ too. Thus in the quotient $p(x, y) = p(0, 0) + x(\cdots) + y(\cdots) = p(0, 0)$ always, and our map is indeed injective, hence bijective, hence an isomorphism.
