Exercise
Consider the following polynomials in Q[X,Y].
Pick out the correct assertion.
h 
(f) + (g)
g (f) + (h)
f (g) + (h)
No. (f) + (g) = ({X - Y, XY - X})
and the morphism Q[X,Y] -> Q
determined by X -> 1, Y -> 1
maps the ideal to {0} but h to 1.
Yes, g = XY - Y = (X - 1)Y
(h).
(h).
No. (g) + (h) = (Y)
and the morphism Q[X,Y] -> Q[X]
determined by X -> X, Y -> 0
maps the ideal to (Y) to {0} but f to -X.