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The following passage from Wikipedia's article on quotient rings gives an example of a quotient ring of polynomials over a finite field.

"One important instance of the previous example is the construction of the finite fields. Consider for instance the field $F_3 = Z/3Z$ with three elements. The polynomial $f(x) = x^2 + 1$ is irreducible over $F_3$ (since it has no root), and we can construct the quotient ring $F_3[x]/(f)$. This is a field with $3^2=9$ elements, denoted by $F_9$. The other finite fields can be constructed in a similar fashion."

I have two questions: (1) Why must the resulting field have $3^2=9$ elements? (2) Why does $f(x)$ having no root imply that it's irreducible?

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    $\begingroup$ 1) It's a vector space over $\mathbb{F}_3$ of dimension $2$, spanned by $1$ and $x$. 2) It's quadratic, so if it were reducible it would have a root. $\endgroup$ Sep 20 '17 at 6:29
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(1) Because in that field, $x^2$ and $-1$ are the same thing. That means that any element has a unique representative with only first-degree and constant term (any higher degree term may be reduced). Each of those places has three possible values (being elements of $\Bbb F_3$), resulting in a total of $3\cdot 3 = 9$ possible elements. Here they are: $$ \begin{array}{ccc} 0x + 0&0x + 1 & 0x+2\\ 1x + 0&1x + 1 & 1x+2\\ 2x + 0&2x + 1 & 2x+2 \end{array} $$ (2) Because if it were reducible, then its factors would be linear. Linear polynomials over fields always have roots. The same argument works for third degree expressions, but not for degree four og higher (the prime example being $(x^2 + 1)^2$).

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Fact 1: Let $k$ be a field, and $f(x)\in k[x]$ be a polynomial of degree $n\geq 1$. Then for any $\overline{g(x)}\in k[x]/(f(x))$, there is a unique polynomial $g_0(x)\in k[x]$ of degree $\leq n-1$ such that $\overline{g(x)}=\overline{g_0(x)}.$

Fact 2: Let $k$ be a field and $f(x)\in k[x]$. If $f(\alpha)=0$ for some $\alpha \in k$, then $f(x)$ is reducible.

(To prove the above two facts, use division algorithm)

(1). In this case, using Fact 1, we see that any element $\overline{g(x)}\in \mathbb{F}_3[x]/(f(x))$ can be written uniquely as $\overline{g_0(x)}$ where $g_0(x)\in \mathbb{F}_3[x]$ is either a constant, or a polynomial of degree one. Since $|\mathbb{F}_3|=3,$ we see that $|\mathbb{F}_3[x]/(f(x))|=3\times 3=9$.

(2). If it has a root, then it is straight forward from Fact 2 that it is reducible. On the other hand, assume it is reducible. The $f(x)=p(x)q(x)$ for some $p(x), q(x) \in \mathbb{F}_3[x]$, and none of them are constants. So $p(x)=ax-\alpha, q(x)=bx-\beta$ for some $a,b,\alpha, \beta \in \mathbb{F}_3$ and $a\neq 0, b\neq 0$. But then both $a^{-1}\alpha$ and $b^{-1}\beta$ are roots of $f(x)$.

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  • $\begingroup$ What does the bar above $g(x)$ mean? $\endgroup$
    – Asker
    Sep 20 '17 at 17:03
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    $\begingroup$ It is the image of $g(x) \in k[x]$ in $k[x]/(f(x))$ under the natural map $k[x] \to k[x]/(f(x))$. $ \overline{g(x)}:= g(x) + (f(x))$. $\endgroup$
    – Krish
    Sep 20 '17 at 17:06

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