Relation of regular map with the map of tangent Cones

Question

I am reading Chapter 1 of J.S. Milne notes. Link is here: https://www.jmilne.org/math/CourseNotes/LEC.pdf

I am confused on Example 2.7(a) on page 18 of the notes. So the situation is we have a regular map $$\phi: \mathbb{A}^{1}\rightarrow V $$ $$t \rightarrow (t^{2}-1,t(t^{2}-1))$$ where $V$ is the variety defined by $Y^{2}-X^{3}-X^{2}=0$.

I guess I am confused on two things. The first one feels obvious as I feel I am just missing some trivial thing. However, for the second question, I feel I am missing something important.

1. Why exactly is the tangent cone to $\mathbb{A}^{1}$ at the point $Q=1$ defined to be $k[s]$ where $s$ is the class of $T-1$ in $m_{Q}/m_{Q}^{2}$?

2. Also, I am having trouble seeing how a regular map $\phi$ induces a map of tangent cones. So my question how does $\phi$ determine that $x$ maps to $2s$ and $y$ maps to $2s$?

Here are my thoughts so far.

For question 1, from definition, if our variety is $V=Spec(k[X_{1},...,X_{n}])/\frak{a}$, then we take the initial part of the polynomials defined in the ideal $\frak{a}$. However, for question 1, isn't affine space $\mathbb{A}^{1}$ defined by the zero polynomial. Of course, the definition is defined for the origin so as $Q=1$, it would be $Spec(k[T-1]/(0))$. But $\frak{a}=0$ still so I feel so I am not sure what the tangent cone is. In other words, initial part of $0$ is $0$ right?

For question 2, I am just not sure how to go from a regular map to maps of the tangent cone. I looked on Tangent Spaces and Morphisms of Affine Varieties which gives me an idea of how regular map induces a map of tangent spaces through the Jacobian. However, I am not sure how a regular map induces a map of tangent cones. I tried playing around with variables to see how to even get $2s=2(T-1)$.

So as $m_{Q}=T-1$ the quotient $m_{Q}/m_{Q}^{2}$ is $(T-1)/(T^{2}-2T+1)$. From the regular map, $x$ maps to $T^{2}-1$ which is $2T-1-1$ which is $2(T-1)=2s$. I am assuming that is how we figure out where $x$ maps to. Now, as $y$ maps to $t(t^{2}-1)$, we do not get the same result. My guess is since on the tangent cone on $V$, $y=x$ or $y=-x$, we get $y$ maps to $2s$ or $-2s$. I feel I am missing something important for question 2.

Answer 1

For (1), note that Milne defines the tangent cone only at the origin in the definition at the top of page 18. In order to apply this specific definition to points that aren't the origin, we have to do some translating. For (1), this means we need to apply the isomorphism $k[s]\cong k[t-1]$ which moves $1$ to the origin, and then calculate with the lowest-degree homogeneous term of $s$, which is $s$. In general, if we want to use the "lowest degree" method at points that aren't the origin, what we should do to avoid translating is to use the lowest-degree terms of the Taylor expansion of our function at that point. In this case, the Taylor expansion of $t-1$ at the point $1$ is just $(t-1)$ and we can use this to make the computation.

An alternate construction of the tangent cone which might help you deal with these issues is that the tangent cone at a point $p$ in a scheme $X$ is given by $\operatorname{Spec} \bigoplus_{i=0}^\infty \mathfrak{m}^i/\mathfrak{m}^{i+1}$ where $\mathfrak{m}$ is the maximal ideal of the local ring $\mathcal{O}_{X,p}$. (This is mentioned slightly further down on page 18.) This also gives you the same answer as the above paragraph: $\mathfrak{m}^i=(T-1)^i=( (T-1)^i)$, so the vector space $\mathfrak{m}^i/\mathfrak{m}^{i+1}$ is just one-dimensional - it's just the span of $(T-1)^i$, so $\bigoplus_{i=0}^\infty \mathfrak{m}^i/\mathfrak{m}^{i+1}\cong k[T-1]$ and things work out.

For (2), recall that a map of affine schemes $f:X\to Y$ sending $x$ to $y$ gives a map $f^*:k[Y]\to k[X]$, which sends $\mathfrak{m}_y\to\mathfrak{m}_x$ as per the linked answer. By the same logic, we get maps $\mathfrak{m}_y^i/\mathfrak{m}_y^{i+1}\to \mathfrak{m}_x^i/\mathfrak{m}_x^{i+1}$ (in fact, this generalizes by replacing the map $f^*:K[Y]\to K[X]$ with the map $f^*:\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$). So we get a map on tangent cones.

Your method of finding where $x$ goes under the map on tangent cones is correct. For $y$, as confirmed in the comments by André 3000, one may compute via polynomial long division that $y$ maps to $t^3-t=(t+2)(t-1)^2+2(t-1)$, which is equal to $2(t-1)$ modulo $(t-1)^2$, and thus $y$ is sent to $2s$. This is essentially the same as what you were doing: you were rewriting the highest degree term in as $t^2$ multiplied by a polynomial in $t$, replacing $t^2$ with $2t-1$, and then repeating.