I am reading the proof of the fact that as a group under multiplication, the set of nonzero elements of (the Galois field/finite field of order ) is isomorphic to and is therefore cyclic.
During the course of the proof, the author reminds the reader that the operation is componentwise addition. I do not see why this is true...we are looking at the multiplicative group of the field. Can someone explain? (I will add more detail if necessary).
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 ."
Also if you are having trouble understanding this I, at the risk of shamelessly self-promoting, suggest you see this post I wrote on my blog about this topic.
I'm sorry for the error, does that make more sense?
It's fine, everyone's entitled to some mistakes (I certainly made some as well). I do see it clearly now. Just one thing... You wrote our map as , but should it really be (specifically from the multiplicative group of our field to the direct product)?