= Solution
The conjugates of $\omega$ are $\omega,ldots,\omega^{p-1}$. From
$$
X^p-1=(X-1)\Phi_p(X)
$$
we obtain, at every nontrivial $p$th root $\alpha$,
$$
\Phi_p'(\alpha)=\frac{p\alpha^{-1}}{\alpha-1}.
$$
The product of the roots is one and $\prod_{a=1}^{p-1}(1-\omega^a)=\Phi_p(1)=p$. The derivative formula for a <polynomial discriminant> therefore yields the <discriminant of a prime cyclotomic power basis>:
$$
\boxed{\operatorname{disc}(1,\omega,\ldots,\omega^{p-2})
=(-1)^{(p-1)/2}p^{p-2}.}
$$
Since $p\equiv1\pmod4$, the sign is positive. The <Vandermonde determinant>
$$
\Delta=\prod_{1\leq a<b\leq p-1}(\omega^a-\omega^b)
$$
belongs to $L$ and satisfies $\Delta^2=p^{p-2}$. Hence
$$
\boxed{\sqrt p=\mathord\pm\frac{\Delta}{p^{(p-3)/2}}\in L.}
$$
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