Elliptic curve: Type of reduction mod 2, how can I show the curve has a cusp?

Question

I want to know what type of reduction the curve $E : y^2 = x^3 + 7x$ has at $p=2$. From online search, I obtain that it has additive/cuspidal reduction. But this disagrees with my own computation, which means I must be doing something wrong. My own computation is this:

Modulo 2, the curve becomes $y^2 = x^3 + x$. This has a double root at $(1, 0)$. So I make the change of coordinates $x' = x-1$ to shift the singular point to $(0, 0)$, and the curve in the new coords (after relabelling $x'$ back to $x$) is $y^2 = x^3 + x^2$. Rearranging, this is $$ x^3 + x^2 - y^2 = 0 $$ This can be viewed as the Taylor expansion of my curve at $(0, 0)$, and so there is a double point, and the tangent lines are given by factorizing $(x^2 - y^2) = (x-y)(x+y)$. From this, I conclude that there is split multiplicative reduction. What am I doing wrong? Does it have something to do with how over $\mathbb{F}_2$, the tangent lines $(x-y)$ and $(x+y)$ are actually the same lines?

I would really appreciate if you could

1. tell me where I am wrong in this 'proof'.
2. show me how to do it correctly.
3. bonus helpfulness if you could direct me a good resource to understand this concept of computing the type of singular point properly/efficiently.

Thank you very much!

Answer 1

Your work is fine through showing that your curve is the same as $y^2=x^3+x^2$, but the conclusion that this has a node in characteristic two is not. Recall that a geometrically, a node should have two distinct tangent lines - your curve does not, since $(x-y)=(x+y)$ in characteristic two. (See this answer for a comparison of some explicit definitions of a node.) It's fairly well-known that when you're dealing with a cubic, it may have at most one singular point, which is either a node or a cusp, and checking the tangent directions as you've done is enough to determine what type of singularity it is.

As far as why this is a cusp, Tabes Bridges has provided the correct solution in the comments: the substitution $y\mapsto x+y$ takes your curve to $y^2+x^2=x^3+x^2$, or $y^2=x^3$, the standard cusp.