The algebraic structure of ring $\mathbb{C}[x^2,xy,y^2]$ $1.$ What are maximal ideals of this ring?
$2.$ Is this ring local?
$3.$ Is this ring regular?
EDIT:
I tried to construct some ring that is isomorphic to $\mathbb{C}[x^2,xy,y^2]$
, but I cannot find it.
In addition, I think ideal $(x^2,xy,y^2)$ is a maximal ideal in $\mathbb{C}[x^2,xy,y^2]$.
 A: I'm afraid I don't know the answer to whether this ring is regular. I have a few thoughts on it at the end of the post and if I come up with something I'll edit this post, but I can answer your first two questions.
(EDIT: Please see the comments for a solution on regularity.)
A more familiar way to write this ring would be $\mathbb C[r, s, t]/(rt - s^2)$. Consider the map $\mathbb C[r, s, t] \longrightarrow \mathbb C[x^2, xy, y^2]$ via $r \mapsto x^2$, $t \mapsto y^2$, $s \mapsto xy$. This is onto and vanishes on $rt - s^2$ so it yields a well defined surjection $\phi: \mathbb C[r, s, t]/(rt - s^2) \longrightarrow \mathbb C[x^2, xy, y^2]$. We now have to prove injectivity. Essentially, the idea is that the only relations $x^2$, $xy$, and $y^2$ satisfy are that $x^2 y^2 = (xy)^2$, so the relation $rt = s^2$ will suffice. Anyway, take some $\overline{f} \in \mathbb C[r, s, t]/(rt - s^2)$ that is in the kernel of this map $\phi$. We have the relation $\overline{r}\overline{t} = \overline{s}^2$ Hence, any $s^2$ term in $f$ reduce to $\overline{r}\overline{t}$. Thus, we can without loss of generality take $f$ to be of the form $f_1(r, t) + f_2(r, t) s$, since we only care about $f$ modulo $(rt - s^2)$. Now, we assumed that $\phi(\overline{f}) = 0$ so $\overline{f}(x^2, xy, y^2) = 0$. Remember that this map was well defined on the quotient ring, so that just means that $f_1(x^2, y^2) + f_2(x^2, y^2) xy = 0$. The question then is when is this possible? Well consider the $x$ terms in both summands of this expression. All $x$-terms in $f_1(x^2, y^2)$ must appear with even degree and all $x$ terms in $f_2(x^2, y^2)xy$ must appear with odd degree. Thus, for their sum to be $0$, we need $f_1(x^2, y^2) = f_2(x^2, y^2) = 0$. Hence, each coefficient of $f_1$ and $f_2$ must be zero. Well since $\overline{f} = \overline{f_1(r, t) + f_2(r, t) s}$, we therefore have that $\overline{f} = 0$ and $\phi$ is injective, hence an isomorphism.
Now, recall the correspondence between ideals of a quotient ring and of the ring itself. The (maximal, prime) ideals of $\mathbb C[r, s, t]/(rt - s^2)$ correspond to the (maximal, prime) ideals of $\mathbb C[r, s, t]$ containing $(rt - s^2)$. We can use a little geometry. As $\mathbb C$ is algebraically closed, the Nullstellensatz tells us that all maximal ideals of $\mathbb C[r, s, t]$ are of the form $(r - a, s - b, t - c)$ (that is, they correspond to points in $\mathbb C^3$). Furthermore, $(r - a, s - b, t - c)$ is the ideal of polynomials vanishing on $(a, b, c)$. Hence, $(rt - s^2) \subseteq (r - a, s - b, t - c)$ iff $rt - s^2 \in (r - a, s - b, t - c)$ iff $ac - b^2 = 0$. Thus, the locus of points in $\mathbb C^3$ satisfying $ac - b^2$ correspond precisely to the maximal ideals of $\mathbb C[r, s, t]/(rt - s^2)$ via $(r - a, s - b, t - c)/(rt - s^2)$. This is a complete description of the maximal ideals of this ring. There are clearly many points on this locus so there are many maximal ideals.
I would also like to point out an interesting geometric note about this ring. An important concept used here is that polynomial rings (and their quotients) acts as functions on affine space (and subsets cut out by polynomials - Zariski closed subsets). This dualism between algebra and geometry is fundamental to, you guessed it, algebraic geometry. So I'd like to point out a seemingly unrelated question. How can we describe $\mathbb C^2 / \pm$, where by $/ \pm$ I mean modding out by $(a, b) \sim (-a, -b)$?. Well as algebraic geometry suggests, let's think about the ring of polynomial functions on $\mathbb C^2$ and the quotient $\mathbb C^2 / \pm$. The former is, of course, $\mathbb C[x, y]$. The latter is the subring $\{f \in \mathbb C[x, y] : f(x, y) = f(-x, -y)\}$. I won't prove this here but it's not too hard to see that this is actually exactly the ring in question - $\mathbb C[x^2, xy, y^2]$.
Well we just showed that $\mathbb C[x^2, xy, y^2] \cong \mathbb C[r, s, t]/(rt - s^2)$ via $r \mapsto x^2$, $s \mapsto xy$, $t \mapsto y^2$. We saw also that $\mathbb C[r, s, t]/(rt - s^2)$ has an intimate relation with the zero set of $rt - s^2$, which I will denote $Z(rt - s^2)$. Explaining this is great detail is the job of an algebraic geometry course, so I'll leave some depth aside and just give you some things to ponder. The ring of polynomials acting on this zero set $Z(rt - s^2)$ is best described as $\mathbb C[r, s, t]/(rt - s^2)$. Since this is isomorphic to $\mathbb C[x^2, xy, y^2]$, which is the ring of polynomials acting on $\mathbb C^2 / \pm$, this suggests a relation between $\mathbb C^2/\pm$ and $Z(rt - s^2)$. Well remarkably these are "isomorphic". I'm really thinking of this as an isomorphism of varieties, but again, that's a discussion for an algebraic geometry class. Anyway, here's the map: $\mathbb C^2/ \pm \longrightarrow Z(rt - s^2)$ via $(a, b) \mapsto (a^2, ab, b^2)$. This is a bijection. There's something remarkable about this too - it's in the opposite direction of the map I described between the rings of polynomials on these two respective spaces but it's still very similar to that map.
All of this rambling was not in the scope of your question, but I promised something about regularity so here it is. If you plot $rt - s^2 = 0$ in $\mathbb R^3$ you get a double cone meeting at the origin. Of course, we should really be in $\mathbb C^3$ but I can't visualize this so this is the best I got. The origin is notably different here - it's the one point that stops this thing from being a manifold. More specifically, you can't define a tangent space at the origin, but you can everywhere else. That signifies that this ring is not regular and that the culprit is at the origin. Via this correspondence between algebra and geometry I keep bringing up, this suggests that you should localize at the prime ideal $(r, s, t)/(rt - s^2)$ and see if what you get is a regular local ring. My suspicion, due to this picture, is that it is not and that every other localization by a maximal ideal will be. If I figure this out rigorously I'll edit the post, but if not I hope this rambling helped.
