Sage: Constructing maximal ideal of a local ring, $M = \langle x – P_x \rangle$ gives invalid result, but $M = \langle x – P_x, y – P_y \rangle$ works

algebraic-geometrycommutative-algebraelliptic-curveslocal-ringsprincipal-ideal-domains

Anothony Knapp's book on Elliptic Curves, page 350 states:

$M$ is principal given by $M = \langle t \rangle$ for any element $t$ of $M$ not in $M^2$.

He later writes on that page, that $t$ is the uniformizer of $C$ and

$f = at^l$ for some unit $a$ in $\bar{k}[C]_x$

Washington's book on Elliptic Curves, page 341 also says:

A natural choice is $u_P = x – x_0$ when $y_0 \neq 0$ and $u_P = y$ when $y_0 = y$.

Both of them are saying that when we have the local ring $\mathcal{O}_P$, which is localized at a point $P$, then its maximal ideal will be $M_P = \langle x – P_x \rangle$ (when $P_y \neq 0$).

Let $E(\mathbb{F}_{11}): y^2 = x^3 + 4x, P = (2, 4)$. Then in sage I try to construct $M_P = \langle x – 2 \rangle$.

sage: R.<x> = FunctionField(GF(11)); _.<Y> = R[]
sage: K.<y> = R.extension(Y^2 - x^3 - 4*x)
sage: o = R.maximal_order()
sage: O = K.maximal_order()
sage: M = O.ideal(x - 2)
sage: M
Ideal (x + 9) of Maximal order of Function field in y defined by y^2 + 10*x^3 + 7*x
sage: (x - 2) in M, (x - 2) in M^2
(True, False)

Just as expected. $(x – 2) \in M_P, (x – 2) \notin M_P^2$.

By simple substitution, we can see that the same curve $E$ can also be written as $(y – 4)(y + 4) = (x – 2)^3 – 5(x – 2)^2 – 6(x – 2)$ showing that $(x – 2)$ is a basis for functions on the curve. We confirm this in sage:

sage: (y - 4)*(y + 4) == (x - 2)^3 - 5*(x - 2)^2 - 6*(x - 2)
True

The problem starts now when we take $f = y – 2x = -2(x – 2) + (y – 4) = -2(x – 2) + \frac{(x – 2)^3 – 5(x – 2)^2 – 6(x – 2)}{(y + 4)} \in M_P = \langle x – 2 \rangle$. This is clearly defined on $E$ so $f \in \langle y^2 – x^3 – 4x \rangle$. Yet sage claims it isn't inside the ideal $M_P$:

sage: y - 2*x in M
False

However if I define $M = \langle x – 2, y – 4 \rangle$, then I get the right answer:

sage: M = O.ideal(x - 2, y - 4)
sage: f = y - 2*x
....: f in M, f in M^2, f in M^3
(True, True, False)

What confuses me even more is if I take the ideal $\langle x – 2 \rangle$ from the original function field and decompose it with the quotiented field, then I can see the ideal in there.

sage: o = R.maximal_order()
sage: M2 = o.ideal(x - 2)
sage: O.decomposition(M2)
[(Ideal (x + 9, y + 4) of Maximal order of Function field in y defined by y^2 + 10*x^3 + 7*x,
  1,
  1),
 (Ideal (x + 9, y + 7) of Maximal order of Function field in y defined by y^2 + 10*x^3 + 7*x,
  1,
  1)]

Can someone explain to me why sage says $f \notin M = \langle x – 2 \rangle$? What am I doing wrong? Why does using $M = \langle x – 2, y – 4 \rangle$ work? Surely $\langle x – 2 \rangle$ should be sufficient since they are maximal as claimed by Knapp and Washington.
Please explain showing where I made a mistake in my thinking.

Thanks

Best Answer

Your ring $O = K.{\tt maximal\_order}()$ is not the local ring ${\cal O}_P$ and your ideal $M = O.{\tt ideal}(x-2)$ is not maximal: there are two points with $x$-coordinate $2$ on the curve, namely $(2,4)$ and $(2,-4)=(2,7)$. So your ideal $M$ is contained in $(x-2,y-4)$ and in $(x-2,y+4)$.

Related Question