How to calculate this Zariski cotangent space explicitly?

Question

$\newcommand{\mp}{\mathfrak{m}_p}$Let $\mathcal{C}=\{ (x,y): x^2+y^2-1=0 \} \subset \mathbb{C}^2$, let $p \in (a,b)$, and let $$\mathfrak{m}_p = \langle x-a,y-b\rangle/\langle x^2 + y^2 -1\rangle. $$ The problem is to show that $\mp/\mp^2$ is a 1-dimensional vector space over $\mathbb{C}$, of the form (I believe) $\{ (ta,tb): t\in \mathbb{C} \}$ (because I am supposed to relate $\mp/\mp^2$ with the span of $\nabla(x^2+y^2-1)(p)$, since they are both supposed to be alternative ways of describing the cotangent space at $p$).

I have already shown that $$x^2+y^2-1 = (x+a)(x-a)+(y+b)(y-b) $$ because $p\in (a,b) \implies a^2 + b^2 -1=0,$ and I have also already shown (see p.2 here) that: $$\mp/\mp^2 = \{ f_1(a,b)(x-a) + f_2(a,b)(y-b): f_1, f_2 \in \mathbb{C}[x,y]/\langle x^2 + y^2 -1 \rangle \} $$ However, this is not of the form I want, and thus not one-dimensional (I think), unless $f_1(a,b)=f_2(a,b)$, and I am completely at a loss for how to show this.

Can anyone tell me what I am doing wrong?

This is Exercise 4.14.8, p. 238 of Algebraic-Geometry: A Problem-Solving Approach by Garrity et al. This isn't homework, I'm just stuck on this problem, which is the second-to-last problem of the section, and I want to move on to the next section.

EDIT: The calculation done here says that the dimension of $\mp/\mp^2$ for an analogous hypersurface is two-dimensional -- in fact, if one replaces $\bar{x}$ with $x-a$ and $\bar{y}$ with $y-b$ everywhere, it looks almost exactly the same as my work calculating $\mp/\mp^2$, except that at the end the author is able to conclude that $\mp/\mp^2 \simeq (k \cdot \bar{x} \oplus k \cdot \bar{y} \oplus \langle \bar{x}^2, \bar{x}\bar{y}, \bar{y}^2 \rangle)/\langle \bar{x}^2, \bar{x}\bar{y}, \bar{y}^2 \rangle \simeq k\cdot x \oplus k \cdot y$ which is supposed to be two-dimensional. In contrast, the authors of my book (4.15.3 on p.240) state that the dimension of any irreducible smooth hypersurface in $\mathbb{C}^n$ is $n-1$, and that the dimension of the tangent space (and thus of the cotangent space) agrees with the dimension at all smooth points. And $x^2 + y^2 -1$ is a non-degenerate conic section, hence smooth. Is the hypersurface defined by $xy - x^6 - y^6$ singular at $(0,0)$? Otherwise I have no idea how to explain the discrepancy.

The formula at the bottom of the first page here is almost certainly relevant -- I am still figuring out how to apply it in this instance.

Answer 1

You write that you have shown that $(x+a) (x-a) + (x+b)(x-b) = x^2 + y^2 - 1.$ This mean, when you work modulo the $x^2+y^2 - 1$, you have shown that $(x+a)(x-a) + (x+b)(x-b) \equiv 0.$ But $\mathfrak m_p/\mathfrak m_p^2$ is annihilated by $\mathfrak m_p,$ so that on this quotient $x$ and $y$ both acts as scalars.

So, as Mohan wrote in comments, you have found a linear dependence between $(x-a)$ and $(y-b)$, and so you're done.

This is just a form of the implicit, or maybe inverse, function theorem. If $p = (a,b)$ is a smooth point of $f = 0$, then the formula $df_{|(a,b)} = 0$ will always yield a non-trivial linear dependence relation between $x-a$ and $y-b$ in $\mathfrak m_p/\mathfrak m_p^2$.

Answer 2

I'm still not sure how to show that $x+a \equiv_{\mp^2} 2a$ or that $y+b \equiv_{\mp^2} 2b$, and I will add that later if I figure it out, but I finally understand why $x-a$ and $y-b$ are linearly dependent over $\mathbb{C}$ (mod $\mp^2$):

$$0 \equiv_{\langle x^2 + y^2 -1 \rangle} x^2(x^2 + y^2 -1) = x^2(x+a)(x-a)+x^2(y+b)(y-b)\equiv_{\mp^2} 2a^3(x-a)+2a^2b(y-b) $$ when $a\not=0$ works, and if $a=0$ then $y \not=0$, because $-1\not=0 \implies (0,0)\not\in\{(x,y):x^2 + y^2 -1 =0 \}$ so we can use instead: $$0 \equiv_{\langle x^2 + y^2 -1 \rangle} y^2(x^2 + y^2 -1) = y^2(x+a)(x-a)+y^2(y+b)(y-b)\equiv_{\mp^2} 2b^2a(x-a)+2b^3(y-b) $$ although if either $a=0$ or $b=0$ then one of the coefficients in this expression is still $0$ -- however, if $a=0$, then $x-a=x+a=x$, so we have that $(x+a)(x-a)=x^2 \equiv_{\mp^2} 0$, likewise if $b=0$ then $(y-b)(y+b)=y^2 \equiv_{\mp^2} 0$, i.e. either $x$ respectively $y$ is nilpotent, so we get $$0 \equiv_{\langle x^2 +y^2 -1 \rangle} y^2(x)(x)+y^2(y+b)(y-b) \quad with\ y^2x \equiv_{\mp^2} b^2x \not=0,\ y^2(y+b)\equiv_{\mp^2} 2b^3 \not=0 $$ $$0 \equiv_{\langle x^2 +y^2 -1 \rangle} x^2(x+a)(x-a)+x^2(y)(y) \quad with\ x^2(x+a) \equiv_{\mp^2} 2a^3 \not=0,\ x^2(y)\equiv_{\mp^2} a^2y \not=0 $$ and thus linear dependence still holds.

At least when $a\not=0\not=b$, we have as a result that $$0 \equiv_{\langle x^2 +y^2 -1 \rangle, \mp^2} 2a(x-a)+2b(y-b) $$ (i.e. divide by $a^2 \not=0$ or by $b^2\not=0$), which obviously has the same form as the tangent space $$\{(x,y): \nabla f\cdot ((x,y)-(a,b))=0 \}=\{(x,y): 2a(x-a)+2b(y-b)=0 \} $$ which is an even stronger resemblance than I had hoped to show originally.