True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

"To see that the multiplicative group

of nonzero elements of

is cyclic, we first note...that it is isomorphic to a direct product of the form

, where each

divides

. So, for any element

in this product, we have

. (Remember, the operation is componentwise addition.)

Thus, the polynomial

has

zeros in

. Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that

. On the other hand, since

has a subgroup isomorphic to

, we also have

. It follows, then, that

is isomorphic to

."