3
$\begingroup$

Let $k$ be a field and $f\in k[a,b]$ an irreducible two-variable polynomial, $B := k[a,b]/(f)$ and $C$ the integral closure of $B$ in its fraction field.

I call $f$ good if it is irreducible in the ring $k[[a,b]]$ of formal power series; equivalently (Nagata, Local Rings, p. 122, Ex. 1), if $C$ has exactly one prime ideal lying above $\mathfrak m = (a,b)B$ (hence, $f$ being good just means that the curve $B$ is analytically irreducible at the origin).

I'm looking for examples of "good" polynomials $f$ such that the residue field $L$ of $C$ is a proper extension of $k$.

I can show (at least if $k$ is infinite) that $[L:k] \cdot r = \mu(f)$, where $\mu(f)$ is the degree of the lowest-degree summand of $f$, and $r$ is the ramification index of $\mathfrak m$ in $C$. Hence, it is clear that $\mu(f)$ must be large enough if one wants interesting examples.

The only "generic" class of examples I could come up with is the one where $f$ is homogeneous: Then $f$ being irreducible implies $f$ is good, and $[L:k] = \deg f$.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

this is the answer_bot. Love your question. I am sure that in the meantime you have moved on and are studying fully faithful exact functors of derived categories of coherent sheaves, but I am still going to answer this one. Yeah!

We can construct examples by starting with a normal affine algebraic curve C over k and a closed point c of C with any given residue field L. If L/k is finite separable, this is always possible even with C being geometrically irreducible and smooth over k. I just made some examples where k has characteristic p > 0 and L is k[x, y]/(x^p - a, y^p - b) which I think generalizes. So there are lot's of L that occur.

Anyway, we next choose a general projection C ---> A^2_k (with coordinates a, b) which maps our chosen point c to (0, 0). The image of C is V(f) for some irreducible f. Since by construction (this is where the "general" above comes in) there is only one point of C above (0, 0) you get an example of what you want.

You can do this explicitly because you can make explicit curves C and then explicitly project and compute the equation f by taking a resultant. Good luck!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .