error: no ambient ring present / computing kernels of maps to QQ

[Andrew Critch]

Hello, 

I'm wondering, what is the intended way to have to Macaulay2 compute the kernel a ring map to QQ?

R=QQ[x]

f=map(QQ,R,{x => 3})

kernel f

produces "error: no ambient ring present".  If I make the target ring QQ[y] instead of QQ, then it works:

R=QQ[x]

f=map(QQ[y],R,{x => 3})

kernel f

produces "ideal(x-3)".  But is a bit unnatural.  So I tried making M2 consider QQ as a polynomial ring by writing QQ[].  M2 accepts this ring, but computes the kernel incorrectly:

R=QQ[x]

f=map(QQ[],R,{x => 3})

kernel f

produces "ideal ()".  So I'm very curious to know how this is supposed to be done.  

Thanks for any input!

--

Critch, Andrew
UC Berkeley
http://math.berkeley.edu/~critch/

[Andrew Critch]

[Grayson, Daniel R.]

Your third example is a bug in Macaulay2. The root cause is that "blocks" in the monomial ordering
that contain no variables are elided from the list of blocks, resulting in unexpected behavior:

i1 : R = QQ[x,y,z, MonomialOrder => {Eliminate 0, Eliminate 1}]

o1 = R

o1 : PolynomialRing

i2 : selectInSubring(1,vars R)

o2 = | y z |

1 2
o2 : Matrix R <--- R

i4 : R = QQ[x,y,z, MonomialOrder => {Eliminate 1, Eliminate 1}]

o4 = R

o4 : PolynomialRing

i5 : selectInSubring(1,vars R)

o5 = | y z |

1 2
o5 : Matrix R <--- R

i6 : R = QQ[x,y,z, MonomialOrder => {Eliminate 2, Eliminate 1}]

o6 = R

o6 : PolynomialRing

i7 : selectInSubring(1,vars R)

o7 = | z |

1 1
o7 : Matrix R <--- R

I think we should fix that by retaining those blocks, rather than by writing special purpose code to compensate for their absence (which we have already had to do once before). Then the answer on o2 above would be | x y z |. Any comments?

Your first example illustrates that M2 is expecting both source and target to by polynomial rings. We should fix that, too.

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[Andrew Critch]

No intelligent comments, except to agree strongly with this:

Your first example illustrates that M2 is expecting both source and target to by polynomial rings.  We should fix that, too.

I have needed to take the kernel of a map from a polynomial ring to a function field on numerous occasions, and if that ends up being fixed along with the map-to-QQ case, then excellent!

Thanks,

--

Critch

[Michael Stillman]

I agree, we should fix this! Besides these unimplemented features, there are also some inefficiencies in kernels of ring maps that should be fixed. I'll put it on my todo list. If anyone wants to help, please say so!

-- Mike

[Grayson, Daniel R.]

On Dec 20, 2011, at 9:20 PM, Andrew Critch wrote:

>
> I have needed to take the kernel of a map from a polynomial ring to a function field on numerous occasions, and if that ends up being fixed along with the map-to-QQ case, then excellent!
>

I think that already works:

i1 : F = frac(QQ[x,y])

o1 = F

o1 : FractionField

i2 : f = map(F,QQ[r,s,t],{x/y,x^2/(y-1),x+y})

2
x x
o2 = map(F,QQ[r, s, t],{-, -----, x + y})
y y - 1

o2 : RingMap F <--- QQ[r, s, t]

i3 : ker f

2 2 2
o3 = ideal(r t + r s - r*s*t + 2r*s - s*t + s)

o3 : Ideal of QQ[r, s, t]