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Let $a,b$ be odd positive integers, and let $x$ be an integer not divisible by $7$ or $13$.

In the field $K = \mathbb{Q(\theta)}$ with $\theta^3 = 7 \cdot 13^2$, we have $$ x^3 - 7 \cdot 13^2 \cdot (7^a 13^b)^3 = (x-7^a 13^b \theta)(x^2 + 7^a 13^bx\theta + 7^{2a} 13^{2b}\theta^2). $$ How can I check whether the two factors on the right are relatively prime in $\mathcal{O}_K$ (the ring of integers of $K$)? Should I take norms?

I know that an integral basis for $K$ is given by $1,\theta,\theta^2/13.$

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up vote 7 down vote accepted

Taking norms is not the best technique, since it can return false negatives. (E.g. in $\mathbb Z[i]$, the elements $2-i $ and $2+i$ are coprime, but they both have norm equal to $5$.)

The simplest approach is to just reduce one of them modulo the other and see what happens.

In your case it makes sense to reduce the second factor modulo the first, since the first is evidently simpler. Since $x \equiv 7^a 13^b \theta \bmod$ the first factor, we find that modulo this first factor, the second factor is equal to $3 \cdot 7^{2 a} 13^{2 b} \theta^2.$

[This is an illustration of the general principle that if $x_0$ is a root of $f(x)$, then $f(x) = (x - x_0)(f'(x_0) +$ terms divisible by $(x- x_0 ))$. In your case, $f'(x_0) = 3 x_0^2$.]

So now you have to ask yourself whether $3 \cdot 7^{2a} 13^{2b}$ has any factor in common with $x - 7^a 13^b \theta$. Since $x$ is coprime to $7$ and $13$ by stipulation, the only common factor could be a divisor of $3$.

Now it makes sense to consider norms, since $3$ is a rational prime, and so it has a factor in common with $x - 7^a 13^b$ precisely if it divides the norm of this element. The norm is equal to $x^3 - 7^{3a+1}13^{3b+2}.$

Modulo $3$ this is equal to $x - 1$, so if $x \not\equiv 1\bmod 3$ then your two factors are coprime, while if $x \equiv 1 \bmod 3$ then they are not coprime.

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Thanks for teaching me this. – Hoffden Feb 6 '12 at 21:05
Dear Hoffden, You're welcome. Regards, – Matt E Feb 7 '12 at 3:13
What an ingenious answer! – awllower Feb 10 '12 at 13:26

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