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**HallsofIvy** All polynomials with coefficients in K, yes. If p(x)= $\displaystyle ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z$, then $\displaystyle f(p(x))= f(a)f(x)^n+ f(b)f(x)^{n-1}+ \cdot\cdot\cdot+ f(y)f(x)+ f(z)$, because f is an automorphism, $\displaystyle = a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z$ because f "fixes" members of K.

IF x is a root of p(x) then $\displaystyle p(x)= 0$ so $\displaystyle f(p(x))= a(f(x))^n+ b(f(x))^{n-1}+ \cdot\cdot\cdot+ yf(x)+ z= 0$. that is, if x is a zero of a polynomial with coefficients in K, f(x) is a zero of that same polynomial: f permutes the zeros.