Geometric interpretation of ramification of prime ideals.

From my perspective, inertia does not really have a geometric flavor. In other words, looking at $\mathbb{Z}[i]$ and its fiber over $(3)$ is hard to picture because it is very far from our usual intuition of topological spaces. If I really need to imagine what is going on, I would say that the fiber has two points (indeed $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{F}_3$ is of dimension 2 over $\mathbb{F}_3$) but these two points lies in an algebraic extension.

Perhaps a better way to have some geometric intuition is to start with more geometric objects than $\mathbb{Z}[i]$.

So let's have a look at $Y=\operatorname{Spec}A$ with $A=\mathbb{R}[y]$ and $X=\operatorname{Spec}B$ with $B=A[x]/(x^2-y)$. Here $Y$ is the vertical line on our usual plane, and $X$ is the parabola $y=x^2$. The map $X\rightarrow Y$ is given by $(x,y)\mapsto y$. Thus the fibers over $y_0$ are all the points of the parabola lying in the horizontal line $y=y_0$.

So here are the different fibers :

• if $y_0>0$ then there are two such points $(\sqrt{y_0},y_0)$ and $(-\sqrt{y_0},y_0)$. Algebraically : the ideal $(y-y_0)$ in $B$ decomposes as $(x-\sqrt{y_0})(x+\sqrt{y_0})$. So $(y-y_0)B$ is split (in particular it is unramified).
• if $y_0=0$, then there is only one point $(0,0)$. Algebraically the ideal $(y-y_0)$ in $B$ decomposes as $(x-x_0)^2$. This is an example of ramification, and as your professor said, you can see that there is a ramification geometrically since then the line $y=0$ is tangent to the parabola.
• if $y_0<0$, then there is no real points, but there is two complex ones : $(i\sqrt{-y_0},y_0)$ and $(i\sqrt{-y_0},y_0)$. These complex points does not comes from two different ideals in $B$ and in fact the ideal $(y-y_0)B$ is prime and the residue field is an extension of degree 2 of $A/(y-y_0)$. So $(y-y_0)$ is inert in $B$.

See, this is not easy to picture it : if we draw only real points, these inert primes should have empty fibers. If we draw complex points, well, $\mathbb{C}$ being algebraically closed, there is no inertia...

Here the map $X\rightarrow Y$ is of degree 2, so a prime in $Y$ is either split, ramified or inert. Of course if the degree is greater than 2, the fiber over each prime may be a bit of everything. For example, look at $X=\operatorname{Spec}A[x]/(y-P(x))$ where $P$ is your favorite polynomial of degree 3 or 4, and draw horizontal lines.

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Now let's have a look at the second picture. I remember that when I first met them, I asked myself questions similar to yours.

• Yes $X$ should be regular, what you see is not $X$.
• Inertia is not represented at all, this is a picture of complex points.
• What looks like a singularity is a representation of a point where locally the map $X\rightarrow Y$ is given by $z\mapsto z^e$ where $e$ is the number of branches going through the point.

The idea of this diagram is to represent the behavior of a solution of an equation $P_{y_0}(x)=0$ where $y_0$ varies.

So take again the parabola $y=x^2$. A diagram of the situation has an X shape. Let $y_0\in Y$ be any point (except 0). Then there is above two solutions to the equation $x^2=y_0$ (remember that we look at complex point and we forgot about inertia). Choose one, say $x_0=\sqrt{y_0}$. Then as $y_0$ varies, $x_0$ varies continuously, until we reach $y_0=0$. Then if $y_0$ keeps moving $x_0$ can choose two different paths. We have an embranchment point. And that is why we may want to draw it with a singularity.

If we want to add inertia, we could imagine points like those of Mumford's picture : inert primes do not make embranchment points, so there should be several line which does not meet through a single point (single because we have only a single prime to represent several points in an algebraic extension). This is what you see above $(3)$ in Mumford's picture for example.