Question: Show that {1, y, , y p- } is a basis for Z p (y) over Z p (y p ), where y is

Show that {1, y,· · ·, yp-¹} is a basis for Zp(y) over Zp(yp), where y is an indeterminate. Referring to Example 51.4, conclude by a degree argument that xp -t is irreducible over Zp(t), where t = yp.

Data from in Example 51.4

Let E = Zp(y), where y is an indeterminate. Let t = yP, and let F be the subfield Zp(t) of
E. (See Fig. 51.5.) Now E = F(y) is algebraic over F, for y is a zero of (xp - t) ∈ F[x]. By Theorem 29.13, irr(y, F) must divide xP -t in F[x]. [Actually, irr(y, F) = xp — t. We leave a proof of this to the exercises (see Exercise 10).] Since F(y) is not equal to F, we must have the degree of irr(y, F) ≥ 2. But note that xP - t = xp - yp = (x - y)p,
since E has characteristic p. Thus y is
a zero of irr(y, F) of multiplicity > 1. Actually, xP — t = irr(y, F), so the multiplicity of y is p.

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