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May 6, 2023 at 20:22 comment added KConrad When that long label is shortened to "inessential discriminant divisor" and we don't know the original context, we mistakenly think "inessential" refers to the prime factors, not the indices, and now the label "inessential" looks wrong because a common prime factor of many numbers looks essential, not inessential. The name "common index divisor" for these primes is much less confusing.
May 6, 2023 at 20:21 comment added KConrad I found the origin of the term "inessential discriminant divisor". Kronecker (p. 313 in eudml.org/doc/148482) was studying integral extensions of $\mathbf C[t]$ and in a polynomial analogue of the formula ${\rm disc}(\mathbf Z[a]) = [\mathcal O_K:\mathbf Z[a]]^2{\rm disc}(K)$, he called the square factor the "inessential divisor" of ${\rm disc}(\mathbf Z[a])$ and ${\rm disc}(K)$ the "essential divisor" of ${\rm disc}(\mathbf Z[a])$. So a prime dividing all indices $[\mathcal O_K:\mathbf Z[a]]$ is a "common prime factor of the inessential divisors of all discriminants." (contd).
May 22, 2016 at 5:29 comment added P Vanchinathan @KConrad: Thanks for providing reasons for studying the full ring of integers of the number field. Your comments have answered the question completely.
May 22, 2016 at 3:37 comment added KConrad The lesson is that you need to give up on the idea of trying to understand everything in a ring of integers of a number field by relying on the crutch of subrings of the form $\mathbf Z[b]$. In fact it was really only after Dedekind discovered the cubic example above that he was compelled to give the full ring of integers of a number field the due respect it deserved, even if it didn't have the form $\mathbf Z[b]$.
May 22, 2016 at 3:33 comment added KConrad The answer to your second question is also no. The (unique) cubic field lying inside the 31st cyclotomic field is $F = \mathbf Q(c)$ where $c^3 + c^2 - 10c - 8 = 0$ (I used PARI to find this). The field $F$ is a Galois extension of $\mathbf Q$ (since it's inside a cyclotomic extension) and once again it turns out that 2 splits completely in $F$ but for every $b \in \mathcal O_F - \mathbf Z$ we have $2 \mid [\mathcal O_F:\mathbf Z[b]]$, so the minimal polynomial of $b$ has discriminant divisible by 2.
May 22, 2016 at 3:20 comment added KConrad The answer to your first question is no (if you mean for $\theta$ to be an algebraic integer). Dedekind discovered the first example of this: if $a^3 - a^2 - 2a - 8 = 0$ and $K = \mathbf Q(a)$ then $2$ splits completely in $K$ (in particular, 2 is unramified in $K$), but $2 \mid [\mathcal O_K:\mathbf Z[b]]$ for all $b \in \mathcal O_K - \mathbf Z$ and therefore $2$ divides the discriminant of the minimal polynomial of $b$, for all such $b$. The term to google is "essential discriminant divisor" (or, amazingly, also "inessential discriminant divisor," a term that never made any sense to me).
May 22, 2016 at 3:04 history asked 352506 CC BY-SA 3.0