F = E = Z 7 in Theorem 22.4. Compute for the indicated evaluation homomorphism. 3

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F = E = Z7 in Theorem 22.4. Compute for the indicated evaluation homomorphism.

 ∅3[(x4 + 2x)(x3 - 3x2 + 3)]


Data from Theorem 22.4

Let F be a subfield of a field E. let a be any element of E, and let x be an indeterminate. The map ∅α : F [x] → E defined by ∅α(a0 + a1x + anxn) = a0 + a1α + ..... +anαfor (a0 + a1x + · · · + an xn) ∈ F[x] is a homomorphism of F[x] into E. Also,∅α(x) = α, and ∅α maps F isomorphically by the identity map; that is, ∅α(a) = a for a ∈ F. The homomorphism ∅α is evaluation at α.


Proof The subfield and mapping diagram in Fig. 22.5 may help us to visualize this situation. The dashed lines indicate an element of the set. The theorem is really an immediate consequence of our definitions of addition and multiplication in F [ x]. The map ∅α is well defined, that is, independent of our representation off (x) ∈ F [ x] as a finite sum a0 + a1x + · · · + an xn) since such a finite sum representing f (x) can be changed only by insertion or deletion of terms 0xi, which does not affect the value of ∅αU(x)). If f(x)=a0+a1x+····+anxn,g(x)=b0+b1x+···+bmxm, and h(x)= f(x) + g(x) =c0+ c1x + · · · · + crxr, then ∅α(f(x) + g(x)) = ∅α(h(x)) =c0+ C1α + · · · + crαr, while  ∅α(f(x))+ ∅α (g(x)) = (a0 + a1α ..... + anαn) + (b+ b1α + ..... bmαm). Since by definition of polynomial addition we have ci = ai + bi, we see that ∅α(f(x) + g(x)) = ∅α(f(x)) + ∅α(g(x)). 


Turning to multiplication, we see that if f(x)g(x) =d0 +d1x + ...... + dsxs, then ∅α(f(x)g(x) = d0 + d1α + ..... +dsαs, while [∅α (f(x))] [∅α (g(x))] = (a0 + a1α + ..... + αnαn) (b+ b1α + .... +bmαm). Since by definition of polynomial multiplication dj = ∑ij=0 aibj-i, we see that ∅α(f(x)g(x)) = [∅αf(x))][∅α(g(x))]. Thus ∅α is a homomorphism. The very definition of ∅α applied to a constant polynomial a ∈ F[x], where a ∈ F. gives ∅α(a) = a, so ∅α maps F isomorphically by the identity map. Again by definition of ∅α, we have ∅α(X) = ∅α(1x) = 1α= α.

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