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In $\mathbb Z_3[x]$, find all polynomials and classify reducible or irreducible for all polynomials of degree less than 4.

Here is what I am thinking

Def of Irreducible Let F be a field. A nonconstant Polynomial $p(x) \in F[x]$ is irreducible if its only divisors are its associates and th nonzero constanct polnomials (units)

polynomicals of degree 1 are irreducible. times those irreducible to find reducible of higher degree .

Listing all polynomials for fun

Deg 0 $$[0], [1], [2]$$

Deg 1 $$ \begin{matrix} [0+x] & [1+x] & [2+x]\\ [0+2x] & [1+2x] & [2+2x] \end{matrix} $$

Deg2 $$ \begin{matrix} [0+x^2] & [1+x^2] & [2+x^2]\\ [0+x+x^2] & [1+x+x^2] & [2+x+x^2]\\ [0+2x+x^2] & [1+2x+x^2] & [2+2x+x^2] \end{matrix} $$ $$ \begin{matrix} [0+2x^2] & [1+2x^2] & [2+2x^2]\\ [0+x+2x^2] & [1+x+2x^2] & [2+x+2x^2]\\ [0+2x+2x^2] & [1+2x+2x^2] & [2+2x+2x^2] \end{matrix} $$ Deg 3 $$ \begin{matrix} [0+x^3] & [1+x^3] & [2+x^3] \\ [0+x+x^3] & [1+x+x^3] & [2+x+x^3]\\ [0+2x+x^3] & [1+2x+x^3] & [2+2x+x^3] \end{matrix} $$

$$ \begin{matrix} [0+x^2+x^3] & [1+x^2+x^3] & [2+x^2+x^3]\\ [0+x+x^2+x^3] & [1+x+x^2+x^3] & [2+x+x^2+x^3]\\ [0+2x+x^2+x^3] & [1+2x+x^2+x^3] & [2+2x+x^2+x^3] \end{matrix} $$

$$ \begin{matrix} [0+2x^2+x^3] & [1+2x^2+x^3] & [2+2x^2+x^3]\\ [0+x+2x^2+x^3] & [1+x+2x^2+x^3] & [2+x+2x^2+x^3]\\ [0+2x+2x^2+x^3] & [1+2x+2x^2+x^3] & [2+2x+2x^2+x^3] \end{matrix} $$
$$ \begin{matrix} [0+2x^3] & [1+2x^3] & [2+2x^3] \\ [0+x+2x^3] & [1+x+2x^3] & [2+x+2x^3]\\ [0+2x+2x^3] & [1+2x+2x^3] & [2+2x+2x^3] \end{matrix} $$

$$ \begin{matrix} [0+x^2+2x^3] & [1+x^2+2x^3] & [2+x^2+2x^3]\\ [0+x+x^2+2x^3] & [1+x+x^2+2x^3] & [2+x+x^2+2x^3]\\ [0+2x+x^2+2x^3] & [1+2x+x^2+2x^3] & [2+2x+x^2+2x^3] \end{matrix} $$

$$ \begin{matrix} [0+2x^2+2x^3] & [1+2x^2+2x^3] & [2+2x^2+2x^3]\\ [0+x+2x^2+2x^3] & [1+x+2x^2+2x^3] & [2+x+2x^2+2x^3]\\ [0+2x+2x^2+2x^3] & [1+2x+2x^2+2x^3] & [2+2x+2x^2+2x^3] \end{matrix} $$

Need to sort to see if its irreducible or reducible ... trying to see a pattern

Deg 1 In terms of irreducible Monics of degree 1 being $[x],[1+x],[2+x]$ $$ \begin{matrix} [x] & [1+x] & [2+x]\\ [2][x] & [2][2+x] & [2][1+x] \end{matrix} $$ Deg 2 In terms of irreducible Monics of degree 2 being $[1+x^2],[2+x+x^2],[2+2x+x^2]$

$$ \begin{matrix} [x][x] & [1+x^2] & [2+x][1+x]\\ [x][x+1] & ([2+x])^2 & [2+x+x^2] & \\ [x][2+x] & ([1+x])^2 & [2+2x+x^2] \end{matrix} $$ $$ \begin{matrix} [2][x][x] & [2][2+x][1+x] & [2][1+x^2] \\ [2][x][2+x] & [2][2+2x+x^2] & [2]([1+x])^2\\ [2][x][1+x] & [2]([1+x])^2 & [2][1+x+x^2] \end{matrix} $$ Deg 3 starting not to be fun listing evrything. So far there are 6 irreducibles of degree smaller than 3. We can combile them to make 16 combinations which are reducible. Also there are 27 monic possble of degree 3. Now that should mean there are 11 irruducible monics of degree 3. Making 14 irreducible monics to of deg 1,2,3. Meaning there are 28 irreducible of deg of def 1,2,3.Add the 2 nonzero constants and that makes 30 irreducible polynomials.

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There is a very nice way of counting monic irreducible polynomials of each degree, using the fact that if a field has $q$ elements, then they are precisely the roots of $X^q-X$. If you call $N_d$ the number of irreducible monics of degree $d$, then there will be, in your case where the base field has three elements, $2N_d$ irreducibles in all, of degree $d$. Note that invertible elements are conventionally not considered to be irreducible.

I’ll call the field with $q$ elements $\Bbb F_q$, and I’ll call the three elements of $\Bbb F_3$ simply $0$, $1$, and $-1$. There are, of course, three irreducible monics of degree one, namely $X$, $X-1$, and $X+1$.

Now for degree $2$, if $\alpha$ is any root of such an irreducible $f$, it generates a (the) quadratic extension $\Bbb F_9$, and of course there’s another root of $f$. Since there are $6$ elements $\alpha$ of $\Bbb F_9$ not in $\Bbb F_3$, they fall into conjugate pairs $(\alpha,\alpha')$, each such pair giving you an irreducible monic $f$. So there are three irreducible monic quadratics. You’ve seen that these are $X^2+1$, $X^2-X-1$, and $X^2+X-1$.

I’ll let you do the same computation for monic irreducible cubics, to show that there are eight of them.

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The total count depends on whether or not you regard the constant polynomial $f(x)=0$ as irreducible (I will not). In this case, I count 30 irreducible polynomials of degree at most 3 in $\mathbb{Z}_3[x]$.

Suppose that $f(x)=ax^3+bx^2+cx+d$ is a polynomial of degree at most 3. If it is not irreducible, then it must factor as $f(x)=(x-\alpha)g(x)$ for some $\alpha\in\mathbb{Z}_3$.

Therefore, $f(x)$ is irreducible if, and only if $\deg f=1$ or $$\begin{cases}f(0)=d&\neq0\\f(1)=a+b+c+d&\neq 0\\f(-1)=-a+b-c+d&\neq0 \end{cases} $$ There are 6 degree 1 polynomials, and otherwise, there are two possibilities, either $a+c=0$ or $a+c\neq 0$.

Case 1: $a+c= 0$.

Then, $b+d\neq 0$ using the condition $f(1)\neq 0$. There are 3 pairs $(a,c)$ satisfying $a+c=0$, and there are 4 pairs $(b,d)$ with $d\neq 0$ and $b+d\neq 0$. This yields a total of 12 irreducible polynomials.

Case 2: $a+c\neq 0$.

Then $a+c=\pm 1$. The condition $f(1)\neq 0$ above implies $b+d\neq -(a+c)$ and the condition $f(-1)\neq 0$ implies $b+d\neq (a+c)$. In either case, we must have $b+d=0$. There are 6 pairs $(a,c)$ with $a+c\neq0$, and 2 pairs $(b,d)$ with $d\neq 0$ and $b+d=0$. This yields 12 more irreducible polynomials.

This yields a count of 30 irreducible polynomials of degree at most 3 in $\mathbb{Z}_3[x]$. I'm not particularly interested in writing them out, but this can be done using the counting method in this post.

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  • $\begingroup$ g(x) not a constant of course as not to be associate. $\endgroup$ – Tiger Blood Oct 27 '15 at 22:03
  • $\begingroup$ @decko well, this doesn't change the count (unless you want only Monic polynomials) $\endgroup$ – David Hill Oct 27 '15 at 22:11

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