# Calculate the discriminant of a basis [Number Theory]

1. Feb 21, 2012

### Firepanda

Question:

The needed proposition and two examples:

This is as far as I have got:

I need to reduce this (I think) so I can represent is as a matrix! Any idea on how to do this?

Thanks

2. Feb 21, 2012

### morphism

$$(\zeta - 1)Dp(\zeta) = p\zeta^{p-1}$$ and then taking norms of both sides?

3. Feb 21, 2012

### Firepanda

I'm unsure on how to calculate $N(p\zeta^{p-1})$

$N(\zeta - 1) = 5$ though.

4. Feb 21, 2012

### morphism

$N(\zeta - 1)$ isn't 5 - not unless p=5.

As for $N(p \zeta^{p-1})$, this is just $N(p)N(\zeta)^{p-1}$, and $N(p)$ and $N(\zeta)$ are really easy to compute.

5. Feb 21, 2012

### Firepanda

Providing $N(\zeta - 1)=p$ (is this easy to prove?) then I also get $N(p)=p^{2}$ and $N(\zeta)=1$

So my overall answer should be p?

6. Feb 21, 2012

### morphism

$N(p)$ should be $p^{p-1}$.

And as for computing $N(\zeta-1)$, you can either do it the determinant way (and do a bunch of row operations), or you could use the differentiation trick you used above except with $t=1$ isntead of $t=\zeta$.

7. Feb 21, 2012

### Firepanda

Hmm when t=1 I get the denominator as 0, am I not using it right?

8. Feb 21, 2012

### morphism

Notice that
$$t^p - 1 = (t-1)(t-\zeta)\cdots(t-\zeta^{p-1})$$ so the t-1 in the denominator gets canceled off.

Edit:
Actually, we don't want to be looking at (t^p-1)/(t-1), rather just at t^p-1. Differentiate both sides of the equation above and plug in t=1.

9. Feb 21, 2012

### Firepanda

Just to be clear you want me to differentiate both sides of $t^p - 1 = (t-1)(t-\zeta)\cdots(t-\zeta^{p-1})$?

Unsure on how to do the RHS without getting messy!

10. Feb 21, 2012

### morphism

Yes, that's what I want you to differentiate. It won't be too messy (especially if you plug in t=1 after you differentiate). :)

11. Feb 21, 2012

### Firepanda

Hmm I'll admit I'm not sure how to go about doing the RHS!

Do I need to? Why can't I use just the differential of the LHS?

12. Feb 21, 2012

### morphism

If you differentiate the RHS and plug in t=1, everything will immediately vanish except for the first term, which becomes
$$(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1}).$$
This really just follows from the product rule of differentiation.

13. Feb 21, 2012

### Firepanda

Ah I see, sorry I've never used the product rule for more than 2 terms before, I get it now!

so $p = (1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})$

i suppose we want to rearrange this?

14. Feb 21, 2012

### morphism

No real rearrangement is necessary. Just notice that the RHS is also $(\zeta - 1)(\zeta^2 - 1) \cdots (\zeta^{p-1}-1)$ (because p is odd), which is just the product of all the conjugates of $\zeta-1$, a.k.a. _____

15. Feb 21, 2012

### Firepanda

Norm!

So the norm is p? Havn't we got a few too many conjugates though?

16. Feb 21, 2012

### morphism

Yes, the norm is p, and no we have the right amount of conjugates! This is because the conjugates of $\zeta$ are $\zeta$ itself and $\zeta^2, \ldots, \zeta^{p-1}$. Try to actually write down a rigorous proof, using minimal polynomials. It should be easy.

17. Feb 21, 2012

### Firepanda

Ok thanks :)

Is N(p) = p^p-1 since we can write the matrix as a matrix with only p along the diagonal?

18. Feb 22, 2012

Yup!