CoCoA:HowTo:Use Modular Numbers: Difference between revisions
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FALSE | FALSE | ||
------------------------------- | ------------------------------- | ||
Moreover this throws an error | Moreover this command throws an error | ||
Use R::=Z/(5); | Use R::=Z/(5); | ||
=== Answer === | === Answer === | ||
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5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-) | 5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-) | ||
If you want to use 5 as a modular number you should use this syntax (similar to C/C++) | If you want to use 5 as a modular number you should either use this syntax (similar to C/C++) | ||
5 % 5; | 5 % 5; | ||
0 % 5 | 0 % 5 | ||
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0 | 0 | ||
------------------------------- | ------------------------------- | ||
The creation of a polynomial ring with no indeterminated has been disabled to highlight this (unexpected?) behaviour. | |||
Revision as of 14:48, 29 November 2005
Modular Numbers
Question
What's wrong in this?
Use R::=Z/(5)[x]; 5x+2y; 2y ------------------------------- 5=0; FALSE -------------------------------
Moreover this command throws an error
Use R::=Z/(5);
Answer
Type(5); INT -------------------------------
5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-) If you want to use 5 as a modular number you should either use this syntax (similar to C/C++)
5 % 5; 0 % 5 -------------------------------
or embed your integer into the polynomial ring
Use R::=Z/(5)[x]; Poly(5); 0 -------------------------------
The creation of a polynomial ring with no indeterminated has been disabled to highlight this (unexpected?) behaviour.
