Let's return to the polynomial () = 3 2 [] and explore why the term conjugate over really does depend on the field under consideration. We'll continue to label its zeros as 1 = 3 2, 2 = 3 2 ( 1 2 + 3 2 ) , 3 = 3 2 ( 1 2 3 2 ). (1) First, recall that () is irreducible over . To what element(s) over are 1, 2, and 3 conjugate? Are all three conjugate to each other over ? (2) Second, verify that in the polynomial ring (3 2)[], we can factor () into the product () = ( 3 2)(2 + 3 2+(3 2)2). Conclude that () = 2 + 3 2+(3 2)2 is irreducible over (3 2)[]. (3) Finally, to what element(s) over (3 2) are 1, 2, and 3 conjugate? Are all three conjugate to each other over (3 2)