MaBloWriMo 20: the group X star

So, where are we? Recall that we are assuming (in order to get a contradiction) that M_n is not prime, and we picked a smallish divisor q (“smallish” meaning q^2 \leq M_n). We then defined the set X as

X = \{ a + b\sqrt 3 \mid a, b \in \mathbb{Z}; 0 \leq a, b < q \}

that is, combinations of 1 and \sqrt 3 where the coefficients are between 0 and q-1. We defined a binary operation on X which works by multiplying and then reducing the coefficients \pmod q. This is not a group, since for example 0 doesn’t have an inverse. But in the last post we saw that we can make a group X^* simply by including only the elements from X that do have inverses.

Recall that \omega = 2 + \sqrt 3 is in X (as long as q is not 2—which it can’t be, since M_n = 2^n - 1 is odd), and we know that \omega \overline{\omega} = 1, so \omega has an inverse and we conclude \omega \in X^*.

You might enjoy figuring out what X^* looks like in the case q = 3. Tomorrow, we’ll start thinking about the implications of our assumption that s_{n-2} is divisible by M_n, and in particular what it means about the order of \omega in X^*.


About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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