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Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529

Dummit and Foote Exercise 2, Section 13.2, page 529 reads as follows:

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2. Let and let . Obtain fields of 4, 8, 9 and 27 elements by adjoining a root of f(x) to a field F where f(x) = g(x) or h(x) and or . Write down the multiplication tables for for the fields with 4 and 9 elements and show that the non-zero elements form a cyclic group.

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In my first attempt at this exercise I took and

So we have and f(x) as above.

The elements of , then, are as follows:** (see attachment)**

,

,

,

and

The multiplication table can then be composed using the following:

That is ... ... ... ... ... (1)

So the multiplication table, composed using (1) is as follows:** (see attachment)**

So far so good (i think?) but when I test whether the 9 non-zero elements in the multiplication table form a cyclic group they do not

For example if you try a cyclic group for the non-zero elements based on we find:

(?? should generate another member of the group but does not!)

and

(?? not an element of the group)

Can anyone help?

Peter

Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529

In , and . In , you have and . So, your multiplication table for the extension of by is misleading since . So, you have:

Since there are only three non-zero elements, is indeed the generator of that cyclic group. Next, for the field of 9 elements, you need to extend by . There will be 8 nonzero elements of that field extension, and they will, indeed, form a cyclic group. Just make sure you remember how addition and multiplication work in the finite fields. It works just like modular arithmetic.

Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 2, page 529

Thanks SlipEternal - again, most helpful

Peter