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

[Ad123q]

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.

 

[DonAntonio]

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