Why is the discriminant of a linear polynomial equal to  1? This is not really significant to what I am trying to do right now but I'm just wondering...
Why is it that the discriminant for the cyclotomic polynomials $\Phi_{1}(x)$ and $\Phi_{2}(x)$ is 1? 
Using WolframAlpha, the discriminant for other linear polynomials is also 1. Why is that?
Thank you.

If $f(x)$ is a polynomial with complex coefficients, and $a_1\ldots,a_n$ are all its roots (with multiplicity) in $\mathbb{C}$, then by definition the discriminant of $f(x)$ is the product of the squares of the differences between any two roots:
$$f_n^{2n-2}\prod_{1\leq i\lt j\leq n} (a_i-a_j)^2,$$
where $f_n$ is the leading coefficient of $f(x)$.
If $f(x)$ is linear, then $n=1$. So there is no way to pick $i$ and $j$, with $1\leq i\lt j\leq 1$. So the question then becomes: 

How much is a product with no factors?

Well, if we want the associativity law to hold in general, then we need the product with no factors to not affect multiplication. Because if we have $x_1\cdot x_2 \cdots x_n$, we want to be able to associate any which way and get the same answer; but one way to associate would be, for example, $(x_1\cdot x_2)()(x_3\cdots x_n)$ (with a product with no factors in the middle, or really anywhere for that matter). So we need the product with no factors to equal $1$ for everything to work out well.
So we define a product with no factors to be equal to $1$. (For the same reason, a sum with no summands is defined to be equal to $0$).
Since the discriminant of a linear polynomial is a product with no factors, and a product with no factors is equal to $1$, the discriminant of a linear polynomial is equal to $1$. 
Since $\Phi_1(x) = x-1$, and $\Phi_2(x) = x+1$, both are linear, so their discriminant is $1$. 
