Singular

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!

[quote="Leon"]Yes, but [code]f(I) [/code]does not work, because I is an ideal in A and the ground ring is B.[/quote]
Not clear what this should mean. You should use reduce in the form
[code]
reduce(f(I),std(J)));
[/code]
unless you know already that J is a Groebner basis.
But in your case std(J)==1, as one can see directly.

Yes, but [code]f(I) [/code]does not work, because I is an ideal in A and the ground ring is B.

[code]size(reduce(f(I),J))[/code]
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) { .... }[/code]

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);[/code]

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).