- #1
mathbbb2
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Construct fields of each of the following orders: [tex]\textbf{(a)}[/tex] 9 [tex]\textbf{(b)}[/tex] 49 [tex]\textbf{(c)}[/tex] 8 [tex]\textbf{(d)}[/tex] 81 (you may exhibit these as [tex]\frac{F[x]}{(f(x))}[/tex] for some [tex]F[/tex] and [tex]f[/tex]).
Relevant Theorems to use:
[tex]\textbf{(1.)}[/tex] Let [tex]f(x)[/tex] be a polynomial in [tex]F[x][/tex]. [tex]\frac{F[x]}{(f(x))}[/tex] is a field iff [tex]f(x)[/tex] is irreducible.
[tex]\textbf{(2.)}[/tex] [tex]F[/tex] is a finite field of order [tex]q[/tex] and let [tex]f(x)[/tex] be a polynomial in [tex]F[x][/tex] of degree [tex]n \geq 1[/tex]. Then [tex]\frac{F[x]}{(f(x))}[/tex] has [tex]q^n[/tex] elements
Relevant Theorems to use:
[tex]\textbf{(1.)}[/tex] Let [tex]f(x)[/tex] be a polynomial in [tex]F[x][/tex]. [tex]\frac{F[x]}{(f(x))}[/tex] is a field iff [tex]f(x)[/tex] is irreducible.
[tex]\textbf{(2.)}[/tex] [tex]F[/tex] is a finite field of order [tex]q[/tex] and let [tex]f(x)[/tex] be a polynomial in [tex]F[x][/tex] of degree [tex]n \geq 1[/tex]. Then [tex]\frac{F[x]}{(f(x))}[/tex] has [tex]q^n[/tex] elements