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Studying for my algebra exam and looking through old exam exercises I came across the following problem

Let $f = X^4 + 1$, $g = X^2 + X - 1 \in \mathbb{F}_3[X]$ and $\alpha = X + \langle f \rangle \in \mathbb{F}_3[X]/\langle f \rangle$.

a) Find a polynomial $h \in \mathbb{F}_3[X]$ such that $f = gh$ and show that $g$ and $h$ are irreducible.

b) What is the size of $\mathbb{F}_3[X]/\langle f \rangle$? Show that $\alpha^2 + \alpha - 1$ is a zero divisor in $\mathbb{F}_3[X]/\langle f \rangle$

I've already solved a and found $\lvert \mathbb{F}_3[X]/\langle f \rangle \rvert = 81$ for part b, but I'm not sure how to show that $\alpha^2 + \alpha - 1$ is a zero divisor

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Have you done part (a)? How might it be helpful in showing that $\alpha^2+\alpha-1$ is a zero divisor? –  Bey Dec 23 '13 at 12:48
    
Yes I've found that $X^4 + 1 = (X^2 + X - 1)(X^2 - X + 2) - 3X + 3$ so $h = X^2 - X + 2$, but I'm unsure of how this helps me in finding a polynomial $k$ such that $(\alpha^2 + \alpha - 1)k = 0$ –  Emil Dec 23 '13 at 12:52
    
Check the answer below for a very strong hint. –  Bey Dec 23 '13 at 12:54

3 Answers 3

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This situation is analogous to considering the quotient ring $\Bbb Z/(a\cdot b)$ where $a$ and $b$ are primes (irreducibles). At the canonical homomorphism $\Bbb Z\to \Bbb Z/(a\cdot b)$ we know that exactly $a\cdot b$ and its multiples go to zero. In particular, neither $a$ or $b$ does, so in other words, in the quotient ring we have $[a],[b]\ne 0$ but $[ab]=0$. (E.g. $2\cdot 3=0 \pmod6$)

Exactly the same happens here: in $\Bbb F_3[X]/(gh)$ we have $[g],\,[h]\ne 0$ but $[gh]=0$.

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Hint: $g(\alpha)h(\alpha)=f(\alpha)$

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To reformulate Berci's answer in more pedagogical terms:

The point of the quotient ring $\mathbb{Z}/\langle 6\rangle$ is that 6 becomes 0. Since 6 factors as $6=2\cdot 3$ in $\mathbb{Z}$, this means that in the quotient $\mathbb{Z}/\langle 6\rangle$, $2$ and $3$ become zero-divisors.

The point of the quotient ring $\mathbb{F}_3[X]/\langle f\rangle$ is that $f$ becomes 0...

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