Is (2X,5) Really an Ideal in Z[X]?

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I have in my notes that (2X,5) is an ideal of Z[X], but I can't see why this can be so.

For example 5+2X is in (2X,5) and 7+X is in Z[X] but then

(5+2X)(7+X) =
= 35+5X+14X+2X^2
= 2X^2+19X+35.

19 is not divisible by 2 and so this element is not in (2X,5), contradicting the "absorbance" property of ideals.
 
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Ad123q said:
I have in my notes that (2X,5) is an ideal of Z[X], but I can't see why this can be so.

For example 5+2X is in (2X,5) and 7+X is in Z[X] but then

(5+2X)(7+X) =
= 35+5X+14X+2X^2
= 2X^2+19X+35.

19 is not divisible by 2 and so this element is not in (2X,5), contradicting the "absorbance" property of ideals.



You seem to believe that any element in (2x,5) must have an even lineal coefficient, but this is wrong: the 5 there can multiply some x-coeff. of some

pol. and added to the even coefficient in the other factor we get an odd coef.

For example, the element 2x\cdot 1 + 5\cdot x = 7x belongs to the ideal...

DonAntonio
 

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