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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.

enter image description here

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I cannot resist quoting Pólya: "Geometry is the art of correct reasoning on incorrect figures." –  Asal Beag Dubh Feb 14 at 14:28

3 Answers 3

up vote 24 down vote accepted

These pictures are of the real points of the curves, as you correctly point out yourself.
In principle they don't make sense over $\mathbb C$, and even less sense over other fields of an arithmetic nature (like $\mathbb Q$ and its finite extensions) or of positive characteristic.
Experimentally, however, it seems that drawing real pictures can be of help to some mathematicians (and I certainly count myself among them).
It is difficult to formally explain the usefulness of real pictures, but experience shows that one learns to retain from them the features that will be preserved over general fields and to dismiss the ones which are relevant only for $\mathbb R$.
In conclusion, I would advise you not to reject these pictures offhand: they are certainly aesthetically pleasing (and I would argue that this is a very serious argument for a mathematician !) and maybe you will discover that you too are among those who derive insight from their contemplation.

To illustrate (!) what I mean, suppose we want to study the complex subvariety $V(\mathbb C)\subset \mathbb C^2$ defined by $x^2+y^2-1=0$ .
We draw a picture of the good old circle $V(\mathbb R)\subset \mathbb R^2$ with the same equation and can see for example that many lines cut it in two points, a few in one point, and many don't cut the circle at all.
The first two features are preserved in the complex case but experience will make us suspect that non intersecting lines are a feature of the non algebraic closedness of $\mathbb R$. This might help forge some intuition for Bézout's theorem,intersection theory, etc.
The real picture will also lead us to believe that $V(\mathbb C)$ has dimension $1$, which is true in the algebraic geometry sense even though as a topological space in the classical sense the variety has dimension two.
This apparent contradiction is due to the fact that the classical point of view relates to the classical topology of $V(\mathbb C)$, inherited from the metric topology of $\mathbb C$ and to the Brouwer-Lebesgue dimension of spaces, whereas algebraic geometers use the Zariski topology and the Krull dimension.

In conclusion, deducing algebraic geometry properties from real pictures is an art rather than a science, but this has a definite appeal to some of us visualists ...

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Dear Georges, thanks for your answer. I have to say that this business about the "real part of curves" is confusing for me. For example in exercise 3.7 Fulton talk of the real part of the curve $(X^2 +Y^2)^2 + 3X^2Y - Y^3$; what does this mean? Thanks. –  user38268 Mar 16 '13 at 10:41
Dear Benja, the modern point of view in algebraic geometry emphasises a functorial point of view for varieties. In your example this means that your curve $C$ associates to any ring $A$ the set $C(A)$ of pairs $(x,y)\in A^2$ satisfying $(x^2+y^2)^2+3x^2y-y^3=0$. The functor under study is thus $A\mapsto C(A)$ and Fulton is talking about $C(\mathbb R)$. An arithmetic question would be to describe the set $C(\mathbb Z)$, a typical diophantine (hard!) problem . –  Georges Elencwajg Mar 16 '13 at 10:54
Thanks for the clarification! –  user38268 Mar 16 '13 at 11:03
Couldn't agree more. –  Piotr Pstrągowski Mar 16 '13 at 11:14
That was a great exposition, Georges. I totally agree with your point of view. –  busman Mar 16 '13 at 14:01

The pictures are intended to illustrate a few ways in which tangents can be placed at a singular point. In that sense, they make perfect sense: they are perfectly accurate in that respect.

PS. You seem to be too demanding on your pictures! there is no need for the field to be ordered in order to make a pictre with axes. In fact, there is no need for it to even be a field, and you can make useful pictures illusrating interesting points and arguments even in the absurdly general case in which you consider two topological spaces $X$ and $Y$ and draw $X\times Y$ in the «obvious» way.

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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.

enter image description here

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