TITLE: Do the pictures in Hartshorne Ex. 1.5.1 make sense?
QUESTION [11 upvotes]: I have done exercise 1 of section 1.5 of Hartshorne and am able to determine that the curves (a),(b),(c) and (d) are respectively those with a tacnode, node, cusp and triple point. Now when I did this exercise in the back of my mind these were curves in $\Bbb{A}^2_k$ with $k = \Bbb{R}$. 


My question is: If $k$ is any arbitrary field of characteristic not equal to $2$ then do these pictures make sense? 


It seems to me that even to begin to label an $X$ - axis and $Y$ - axis assumes that we have an ordering on the field $k$. What happens if $k = \Bbb{C}$? Surely the pictures will not be accurate anymore.
Here is the relevant picture from Hartshorne.

REPLY [6 votes]: You can visualize these curves also over $\mathbb{C}$. They live in $\mathbb{C}^2 \cong \mathbb{R}^4 \cong \mathbb{R}^3 \times \mathbb{R}$, the usual euclidean space with an additional time parameter. For example, look at the cusp $y^4+y^2 + x^4-x^3=0$ for $(x,y) \in \mathbb{C}^2$. Write $x=a+ib$ and $y=c+it$ with $(a,b,c) \in \mathbb{R}^3$ and $t \in \mathbb{R}$ (the time parameter). The real part of the equation becomes
$a^4 - a^3 - 6 \cdot a^2 \cdot b^2 + 3 \cdot a \cdot b^2 + b^4 + c^4 + c^2 \cdot (1 - 6 \cdot t^2) + t^4 - t^2$
and the imaginary part becomes
$4 \cdot a^3 \cdot b - 3 \cdot a^2 \cdot b - 4 \cdot a \cdot b^3 + b^3 + 2 \cdot c \cdot t \cdot (2 \cdot c^2 - 2 \cdot t^2 + 1)$
For every fixed time $t$ the (complex) curve is an intersection of two surfaces in $\mathbb{R}^3$. One may ask how smoothly these surfaces change with the time. For $t=0$ the intersection with the $ac$-plane gives the real picture from Hartshorne, but there is another component.