I'm attempting the following exercise.
Let $K$ be a field, and let $R$ be the ring $K[X, Y] /\left\langle X^{3}-Y^{2}\right\rangle .$ For any element $a$ of $K,$ show that the ideal
$$
\mathfrak{m}_{a}=\left\langle X-a^{2}, Y-a^{3}\right\rangle
$$
of $R$ is a maximal ideal, with $R / \mathfrak{m}_{a}$ isomorphic to $K$. Moreover, show that if $I$ is the ideal of $R$ generated by $X$ and $Y,$ then $I / m_{a} I$ is a two-dimensional $K$ -vector space for $a=0$.
Here is my proof attempt:
Proof: Let $\phi: R \to K$ such that $\phi$ sends $X,Y,k$ to $a^2,a^3,k$ for $k \in K$. This is a ring homomorphism, and also a surjection. I claim that $\ker(\phi) = \left\langle X-a^{2}, Y-a^{3}\right\rangle$; an element of $\left\langle X-a^{2}, Y-a^{3}\right\rangle$ looks like $P_1 (X-a^{2}) + P_2(Y-a^{3})$ where $P_1, P_2 \in R$, and it is clear that $\ker(\phi)\supset \left\langle X-a^{2}, Y-a^{3}\right\rangle$. Now, let $P(X,Y) \in R$ such that $\phi(P) = P(a^2,a^3) = 0$. Now, note that $P(X,Y)$ can be re-written as a finite sum of variables of the form $(X-a^2),(Y-a^3),k' $ for a constant $k'$, i.e. $P(X,Y) = P(X-a^2+a^2,Y-a^3+a^3) = Q(X-a^2,Y-a^3)$. $ P(a^2,a^3) = 0$ implies that the constant factor of $Q$, $Q(0,0)$ is $0$. Thus, $P(X,Y) = Q(X-a^2,Y-a^3) \in \left\langle X-a^{2}, Y-a^{3}\right\rangle$. By the first isomorphism theorem, we have that $R / \mathfrak{m}_{a} \cong K$.
Here is the that bit I'm even more unsure about: Let $a=0$. Then, $I = \mathfrak{m}_0 = \left\langle X, Y\right\rangle$, and we simply calculate $I/II$. $II = \left\langle X, Y\right\rangle\left\langle X, Y\right\rangle$ is the ideal generated by the finite sum $\sum_{i=1}^n P_iQ_i$ where $P_i,Q_i \in \left\langle X, Y\right\rangle$ are polynomials with degree greater than $0$, with no constant term, in $R$. So every monomial of any element of $II$ will have degree $2$; in fact it is easy to see that any monomial with degree greater than $1$ is an element of $II$. Thus $I/II$ is composed of monomials of degree $1$ in $R$, of which $X,Y$ is a $K$-basis. So $I/II$ is two-dimensional.
Is this a valid proof?
Best Answer
Honestly, the solution is a bit hard to read. Paragraphs would certainly help. As for the bit you are unsure about, the ideas are right but the conclusions are not obvious to me, I'd be more explicit (and what follows is my attempt at doing so).
As you say, the map $e \colon k[x,y] \to k$ mapping $x$ to $a^2$ and $y$ to $a^3$ is surjective. I claim that $\ker e = \langle x-a^2,y-a^3\rangle$. Indeed, take $g$ such that $e(g) = 0$. Since $y-a^3$ is a degree $1$ monic polynomial in $k[x][y] = k[x,y]$, we can write $$g(x,y) = (y-a^3)q(x,y) + r(x)$$ and now dividing $r$ by $x-a^2$ in $k[x]$ gives $$g = (y-a^3)q+(x-a^2)q'+r''.$$ Applying $e$ shows that $r'' = 0$, hence $g \in \langle x-a^2,y-a^3\rangle$, the other inclusion is a direct computation.
Thus $k[x,y]/\langle x-a^2,y-a^2\rangle \simeq k$. Since $\mathfrak m_a$ is the projection of the ideal we're dividing by in $R$, by the third isomorphism theorem we see that
$$R/\mathfrak m_a \simeq k[x,y]/\langle x-a^2,y-a^2\rangle \simeq k,$$
as desired.
The geometric perspective is that $R = k[x,y]/(x^3-y^2)$ is the coordinate ring of the variety in $\Bbb A^2 = k^2$ defined by the equation $x^3 = y^2$. This is a curve, as the image shows:
Points on a variety are associated to certain maximal ideals of its coordinate ring. In fact, when $k$ is algeraically closed there is a bijective correspondence: this is Hilbert's Nullstelensatz. Since $(a^2,a^3)$ satisfies $(a^2)^3 = (a^3)^2$, it is a point lying on the curve, and its associated maximal ideal is $\mathfrak m_a$.
Now, you may notice that the curve is spikey at $(0,0)$. Formally, the curve at that point is not smooth. I will not get into the details, but one way to measure the smoothness of a curve at a point is by calculating the dimension of its tangent space, and checking whether it coincides with the dimension of the variety.
This may be very familiar to us in the differential setting: if $M$ is a smooth $n$-dimensional manifold, at each point the tangent space is an $n$-dimensional vector space.
Our varierty is a curve, this means that its dimension is $1$ (I won't formalize this). Hence, to see that it is not smooth at the point $(0,0)$, we should see that the tangent space is not $1$-dimensional. The tangent space at a point $p$ is defined as the dual of $$ \mathfrak m_p/\mathfrak m_p^2 $$ where $\mathfrak m_p$ is the maximal ideal associated to this point (I don't want to get into this as it is not directly related to your question, but this seemingly contrived definition has geometric meaning). My point is that in this case the tangent space is the dual of $$ \mathfrak m_0/\mathfrak m_0, \qquad \mathfrak m_0 = \langle x,y\rangle $$ and what you are being asked to do is to prove that it has dimension $2$. This is an algebraic way of detecting the spikeyness of the drawing at $(0,0)$.
As for the actual computations, indeed: an element of $\langle x,y\rangle^2$ is a sum of products of the form $QP$ with $Q = xq+yq', P = xp+yp'$. Hence $QP \in \langle X^2, Y^2, XY\rangle$. Since $x^2,y^2,xy \in \mathfrak m_0^2$, we have the other inclusion and $\mathfrak m_0^2 = \langle x^2,y^2,xy\rangle$.
As you point out, this tells us that every element of $\mathfrak m_0/\mathfrak m_0^2$ is of the form $[a+bx+cy]$ for some $a,b,c \in k$. Hence $\{1,x,y\}$ generates the quotient as a $k$-vector space. Linear independence stems from the fact that if $[a+bx+cy] = 0$, then $a+bx+cy \in \mathfrak m_0^2 = \langle x^2,y^2,xy\rangle$ and there exist $p,q,r$ for which
$$ a+bx+cy = x^2 p + y^2 q + xy r. $$
By degree considerations, this forces us to have $p = q = r = a = b = c = 0$.