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image of a ring map
https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=2252		 Page 1 of 1
Author:  Leon [ Thu May 02, 2013 7:03 am ]
Post subject:  image of a ring map
Given algebras A=K[x1,...,xm]/I and B=K[y1,...,yn]/J and homomorphism f:K[x]-->K[y], I'd like to check if f(I)⊆J, which would mean that f defines a homomorphism f: A-->B.

The code below returns an error, since f(I) is undefined. If p∈K[x], how can I calculate f(p)∈K[y]? I've tried subst(p,...); and imap(...), but without success.

Code:
LIB"algebra.lib";
ring A=0,(x,y,z,w),dp;   ideal I=xy+zw,x+y+w+z,x2-w2;
ring B=0,(s,t),dp;   ideal J=s2t2,st+1;
ideal ff=s4,s3t,st3,t4;   map f=A,ff;   reduce(f(I),J);


It would be desirable to be able to check f(I)⊆J automatically, i.e. at once (with a single command) and not manually (for each f(g_i) where g_i are generators of I).
Author:  hannes [ Thu May 02, 2013 11:23 am ]
Post subject:  Re: image of a ring map
Code:
size(reduce(f(I),J))

gives the numnber of non-zero entries, i.e. to test if F(I0 is in J:
Code:
if (size(reduce(f(I),J))==0) { .... }
Author:  Leon [ Thu May 02, 2013 11:58 am ]
Post subject:  Re: image of a ring map
Yes, but
Code:
f(I)
does not work, because I is an ideal in A and the ground ring is B.
Author:  gorzel [ Thu May 02, 2013 1:39 pm ]
Post subject:  Re: image of a ring map
Leon wrote:
Yes, but
Code:
f(I)
does not work, because I is an ideal in A and the ground ring is B.

Not clear what this should mean. You should use reduce in the form
Code:
reduce(f(I),std(J)));

unless you know already that J is a Groebner basis.
But in your case std(J)==1, as one can see directly.
Author:  Leon [ Fri May 03, 2013 12:06 pm ]
Post subject:  Re: image of a ring map
Yes, of course I meant std(J) inside reduce.

Hmm, this is really strange, now my initial code works. I don't know what I made wrong before. Singular always returned an error, saying that I is not defined, since the basering was B and I was in A. Very strange indeed.

Thank you!
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