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let $g(x) = x^2+x-1$ and let $h(x) = x^3-x+1$ obtain fields $4$, $8$, $9$, and $27$ elements by adjoining a root of $f(x)$ to the field $F$ where $f(x)=g(x)$ or $h(x)$ and $F = F_2$ or $F_3$. The teacher gave me the solution to this problem but i do not understand how $F_2[x]/(g(x))$ is a field with four elements, $F_2[x]/(h(x))$ is a field with eight elements, and so on. My question how do you determine or what is the procedure to find how many elements that the field will have. Anyone please help me. All i understand in this problem is that $g(x)$ and $h(x)$ are irreducible in both $F_2$ and $F_3$ thus $F_2[x]/(h(x))$, $F_2[x]/(g(x))$, $F_3[x]/(g(x))$, $F_3[x]/(h(x))$ are fields. How many elements they contain i do not know how to find that.

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  • $\begingroup$ To determine the number of elements in the field do you per harps take the size of the field raise it to the degree of the polynomial for instance we know that the $F_2$ has two elements and degree(g(x))=2 thus $F_2[x]/(g(x)) = 2^2 = 4$. that is how $F_2[x]/(g(x))$ has four element is that true $\endgroup$ – user146269 Apr 30 '15 at 15:29
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For any $p$ prime and $g(x)\in F_p[x]$ then :

$$\# (\mathbb{F}_p[x]/(g(x)))=p^{deg(g)} $$

To do this you have to show that $(1,x,...,x^{deg(g)-1})$ is a $\mathbb{F}_p$-base of the vectorial space $\mathbb{F}_p[x]/(g(x))$. (This works wether $g$ is irreducible or not).

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