Why is the discriminant of a non separable finite extension of a field $0$? I proved for the case when the extension is primitive but can't prove it for the general case. Can trying to use induction help?
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1What is your definition of discriminant? – Kenny Lau Mar 22 '21 at 17:10
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Let $L/K$ be non-separable. By Stacks 030K, the set of elements in $L$ that are separable over $K$ form an intermediate extension $L^s/K$ that is separable, and $L/L^s$ is purely inseparable (Stacks 09HE).
It then suffices to address the case where $L/K$ is purely inseparable (by letting $K := L^s$), by this formula about discriminants and towers of fields.
By Stacks 09HI, one can factorize $L/K$ into a tower $L = E_n / \cdots / E_2 / E_1 / E_0 = K$ where each $E_{i+1}/E_i$ is degree $p$ and purely inseparable and is formed by adjoining the $p$th root of some element in $E_i$. By the same formula quoted above, it suffices to address the case where $L/K$ is one of these steps in the tower.
So let $L = K(a^{1/p})$ for some $a \in K \setminus K^p$. The minimal polynomial of $a^{1/p}$ over $K$ is $f(x) = x^p - a$, so the discriminant of $L/K$ is $f'(a^{1/p}) = p \cdot a^{(p-1)/p} = {\fbox{0}}$, as required. Note that this last step depends on your definition of discriminant.
Kenny Lau
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$Disc(a_1,\ldots,a_m)=\det(A^\top A)=0$ for any $K$-basis of $L$ because $A_{ij}=\sigma_j(a_i)$ where each $\sigma_j\in Hom_K(L,\overline{L})$ is repeated $[L:L^s]$ times so that $\det(A)=0$. – reuns Mar 22 '21 at 17:24
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Hi, so my definition of discriminant is the discriminant of a particular set of basis elements, which is what you have seemed to use. Thank you for the answer. – Yashi Jain Mar 22 '21 at 19:44