Computing extensions of an ideal in Singular or Macaulay2
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Does Macaulay2 or Singular compute extensions of ideals under ring homomorphisms?
Specifically, if $phi : R to S$ is a ring homomorphism (say polynomial rings over $mathbb{Q}$ which can be specified in Macaulay2 or Singular) and $I$ is an ideal in $R$ given by generators, is there a command to compute the ideal generated by ${phi(I)}$ in $S$?
abstract-algebra computer-algebra-systems symbolic-computation computational-algebra macaulay2
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
$endgroup$
add a comment |
$begingroup$
Does Macaulay2 or Singular compute extensions of ideals under ring homomorphisms?
Specifically, if $phi : R to S$ is a ring homomorphism (say polynomial rings over $mathbb{Q}$ which can be specified in Macaulay2 or Singular) and $I$ is an ideal in $R$ given by generators, is there a command to compute the ideal generated by ${phi(I)}$ in $S$?
abstract-algebra computer-algebra-systems symbolic-computation computational-algebra macaulay2
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
$endgroup$
add a comment |
$begingroup$
Does Macaulay2 or Singular compute extensions of ideals under ring homomorphisms?
Specifically, if $phi : R to S$ is a ring homomorphism (say polynomial rings over $mathbb{Q}$ which can be specified in Macaulay2 or Singular) and $I$ is an ideal in $R$ given by generators, is there a command to compute the ideal generated by ${phi(I)}$ in $S$?
abstract-algebra computer-algebra-systems symbolic-computation computational-algebra macaulay2
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
$endgroup$
Does Macaulay2 or Singular compute extensions of ideals under ring homomorphisms?
Specifically, if $phi : R to S$ is a ring homomorphism (say polynomial rings over $mathbb{Q}$ which can be specified in Macaulay2 or Singular) and $I$ is an ideal in $R$ given by generators, is there a command to compute the ideal generated by ${phi(I)}$ in $S$?
abstract-algebra computer-algebra-systems symbolic-computation computational-algebra macaulay2
abstract-algebra computer-algebra-systems symbolic-computation computational-algebra macaulay2
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
edited Dec 8 '18 at 13:47
Rodrigo de Azevedo
13k41958
13k41958
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
asked Nov 4 '15 at 2:31
BabaiBabai
2,62421540
2,62421540
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this is very easy in Macaulay2. See the code exampe below.
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x^2-y,y+2)
2
o2 = ideal (x - y, y + 2)
o2 : Ideal of R
i3 : S = QQ[s,t]
o3 = S
o3 : PolynomialRing
i5 : f = map(S,R,{s^2,t^2})
2 2
o5 = map(S,R,{s , t })
o5 : RingMap S <--- R
i6 : f I
4 2 2
o6 = ideal (s - t , t + 2)
o6 : Ideal of S
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this is very easy in Macaulay2. See the code exampe below.
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x^2-y,y+2)
2
o2 = ideal (x - y, y + 2)
o2 : Ideal of R
i3 : S = QQ[s,t]
o3 = S
o3 : PolynomialRing
i5 : f = map(S,R,{s^2,t^2})
2 2
o5 = map(S,R,{s , t })
o5 : RingMap S <--- R
i6 : f I
4 2 2
o6 = ideal (s - t , t + 2)
o6 : Ideal of S
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
add a comment |
$begingroup$
Yes, this is very easy in Macaulay2. See the code exampe below.
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x^2-y,y+2)
2
o2 = ideal (x - y, y + 2)
o2 : Ideal of R
i3 : S = QQ[s,t]
o3 = S
o3 : PolynomialRing
i5 : f = map(S,R,{s^2,t^2})
2 2
o5 = map(S,R,{s , t })
o5 : RingMap S <--- R
i6 : f I
4 2 2
o6 = ideal (s - t , t + 2)
o6 : Ideal of S
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
add a comment |
$begingroup$
Yes, this is very easy in Macaulay2. See the code exampe below.
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x^2-y,y+2)
2
o2 = ideal (x - y, y + 2)
o2 : Ideal of R
i3 : S = QQ[s,t]
o3 = S
o3 : PolynomialRing
i5 : f = map(S,R,{s^2,t^2})
2 2
o5 = map(S,R,{s , t })
o5 : RingMap S <--- R
i6 : f I
4 2 2
o6 = ideal (s - t , t + 2)
o6 : Ideal of S
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
Yes, this is very easy in Macaulay2. See the code exampe below.
i1 : R = QQ[x,y]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x^2-y,y+2)
2
o2 = ideal (x - y, y + 2)
o2 : Ideal of R
i3 : S = QQ[s,t]
o3 = S
o3 : PolynomialRing
i5 : f = map(S,R,{s^2,t^2})
2 2
o5 = map(S,R,{s , t })
o5 : RingMap S <--- R
i6 : f I
4 2 2
o6 = ideal (s - t , t + 2)
o6 : Ideal of S
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
answered Nov 4 '15 at 7:11
Fredrik MeyerFredrik Meyer
15.3k24165
15.3k24165
add a comment |
add a comment |
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