# Find a zero divisor in the quotient ring $\Bbb{Q}[X]/\langle X^4 - 5X^2 + 6\rangle$ or else prove that none exist.

Find a zero divisor in the quotient ring $\Bbb{Q}[X]/\langle X^4 - 5X^2 + 6\rangle$ or else prove that none exist.

I don't even really know how to start with this. How does one go about finding a zero divisor in a quotient group, or proving that there are none?

Hint: $Q[X]/I$ is integral (i.e. does not contain any zero divisor) if and only if $I$ is prime. Since $Q[X]$ is a UFD, this is equivalent to say that your polynomial is irreducible.

Hint: Try factoring the polynomial $X^4 - 5X^2+6$.

We have

$X^4 - 5X^2 +6 = (X^2 -2)(X^2 - 3) \tag{1}$

in $\Bbb Q[x]$; if we denote the ideal $\langle X^4 - 5X^2 +6 \rangle \subset \Bbb Q[x]$ by $J$, this implies that

$((X^2 - 2) + J)((X^2 - 3) + J) = 0 \tag{2}$

in $\Bbb Q[x] / J$, since

$(X^2 - 2)(X^2 - 3) = X^4- 5X^2 + 6 \in J. \tag{3}$

But

$(X^2 - 2) + J \ne 0 \ne (X^2 - 3) + J \tag{4}$

in $\Bbb Q[x]/J$, since

$\deg X^2 - 2 = \deg X^2 - 3 = 2, \tag{5}$

prohibiting

$X^2 - 2, X^2 - 3 \in J, \tag{6}$

for neither can then be a multiple of $X^4 - 5X^2 + 6$ in $\Bbb Q[x]$. Thus both $X^2 - 2$ and $X^2 - 3$ are zero divisors in $\Bbb Q[x] / J$

In general, on can seek out zero divisors in $\Bbb F[x]/\langle p(x) \rangle$ where $p(x) \in \Bbb F[x]$ for any field $\Bbb F$ by looking for non-trivial factors of $p(x)$ in $\Bbb F[x]$. If $p(x) = q_1(x)q_2(x) \in \Bbb F[x]$, where $\deg q_1, \deg q_2 \ge 1$, then

$(q_1(x) + \langle p(x) \rangle)(q_2(x) + \langle p(x) \rangle) = 0 \tag{7}$

in $F[x] / \langle p(x) \rangle$, so the $q_i(x) + \langle p(x) \rangle$ are zero divisors in $\Bbb F[x]/\langle p(x) \rangle$; if $p(x)$ is irreducible in $\Bbb F[x]$, then $\Bbb F[x]/\langle p(x) \rangle$ is a field and hence no $q(x) + \langle p(x) \rangle \mid 0$ in $\Bbb F[x]/\langle p(x) \rangle$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!