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Let \(K, F\) be number fields with \(F \subseteq K\). Let \(A\) be an additive subgroup of \(K\). Let
and
The relative different \(\mathfrak {d}_{K/F}\) of \(K/F\) is then defined as \(((\mathcal{O}_K)^*)^{-1}\) which one checks is an integral ideal in \(\mathcal{O}_K\). This is also the annihilator of \(\Omega ^1_{\mathcal{O}_K/\mathcal{O}_F}\) if we want to be fancy.
Let \(A,K\) be commutative rings with \(K\) and \(A\)-algebra. let \(B=\{ b_1,\dots ,b_n\} \) be a set of elements in \(K\). The discriminant of \(B\) is defined as
Let \(u \in F\) with \(F=\mathbb {Q}(\zeta _p)\) be a unit such that
for some \(a \in \mathcal{O}_F\). The either \(u = \epsilon ^p\) for some \(\epsilon \in \mathcal{O}_F^\times \) or \(p\) divides the class number of \(F\).
Let \(K/F\) be an extension of number fields and let \(S\) denote the set of \(\alpha \in \mathcal{O}_K\) be such that \(K=F(\alpha )\). Then
where \(m_\alpha \) is denotes the minimal polynomial of \(\alpha \).
Let \(K/F\) be an extension of number fields with \(\mathfrak {p}_F, \mathfrak {p}_K\) prime ideas in \(K,F\) respectively, with \(\mathfrak {p}_K^e \parallel \mathfrak {p}_F\) for \(e {\gt}0\). Then \(\mathfrak {p}_K^{e-1} \mid \mathfrak {d}_{K/F}\) and \(\mathfrak {p}_K^{e} \mid \mathfrak {d}_{K/F}\) iff \(\mathfrak {p}_F \mid e\).
Let \(p\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\) with \(\alpha =\sum _i a_i \zeta _p^i\). Let us suppose that there is \(i\) such that \(a_i = 0\). If \(n\) is an integer that divides \(\alpha \) in \(\mathbb {Z}[\zeta _p]\), then \(n\) divides each \(a_i\).
Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \( \alpha \in \mathbb {Z}[\zeta _p]\). Let \(x\) and \(y\) be integers such that \(x+y\zeta _p^i=u \alpha ^p\) with \(u \in \mathbb {Z}[\zeta _p]^\times \) and \(\alpha \in \mathbb {Z}[\zeta _p]\). Then there is an integer \(k\) such that
Let \(p \geq 5\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(x, y \in \mathbb {Z}\) coprime.
For \(i \neq j\) with \(i,j \in {0,\dots ,p-1}\) we can write
with \(u\) a unit in \(\mathbb {Z}[\zeta _p]\). From this it follows that the ideals
are pairwise coprime.
Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\) and that \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\). Then each of \((x+\zeta _p^i y)\) is divisible by \((1-\zeta _p)\) and there is a unique \(i_0\) such that \((x+\zeta _p^{i_0})\) is divisible by \((1-\zeta _p)^2\).
Let \(p\) be a regular odd prime, \(x,y,z,\epsilon \in \mathbb {Z}[\zeta _p]\), \(\epsilon \) a unit, and \(n \in \mathbb {Z}_{\geq 1}\). Assume \(x,y,z\) are coprime to \((1-\zeta _p)\), \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) and \(x+y\) is divisible by \(\mathfrak {p}^2\) and \(x+ \zeta _p^ky\) is only divisible by \(\mathfrak {p}=(1-\zeta _p)\) (for \(0 {\lt} k \leq p-1\)). Let \(\mathfrak {m}= \mathrm{gcd}((x),(y))\). Then:
We can write
\[ (x+y)=\mathfrak {p}^{p(m-1)+1}\mathfrak {m}\mathfrak {c}_0 \]and
\[ (x+\zeta _p^k y)=\mathfrak {p}\mathfrak {m}\mathfrak {c}_k \](for \(0 {\lt} k \leq p-1\)) where \(m=n(p-1)\) and with \(\mathfrak {c}_i\) pairwise coprime.
Each \(\mathfrak {c}_k=\mathfrak {a}_k^p\) and \(\mathfrak {a}_k\mathfrak {a}_0^{-1}\) is principal (as a fractional ideal).
Let \(K/F\) be a Galois extension of \(F=\mathbb {Q}(\zeta _p)\) with Galois group \(\operatorname{Gal}(K/F)\) cyclic with generator \(\sigma \). Then there exists a unit \(\eta \in \mathcal{O}_K\) such that \(N_{K/F}(\eta )=1\) but does not have the form \(\epsilon /\sigma (\epsilon )\) for any unit \(\epsilon \in \mathcal{O}_K\).
Let \(K\) be a number field, \(p\) be a rational prime below \(\mathfrak {p}\) and let \(\alpha _0, \xi \) be in \(\mathcal{O}_\mathfrak {p}^\times \) (the units of the ring of integers of the completion of \(K\) at \(\mathfrak {p}\)) be such that
where \(\mathfrak {p}^m \parallel p\) and \(r(p-1){\gt}m\). Then there exists an \(\alpha \in \mathcal{O}_{\mathfrak {p}}^\times \) such that \(\alpha ^p =\xi \).
Let \(F=\mathbb {Q}(\zeta _p)\) with \(\zeta _p\) a primitive \(p\)-th root of unity. If \(\xi \in \mathcal{O}_F\) and coprime to \(\lambda _p:=1-\zeta _p\) and if
for some \(\alpha _0 \in \mathcal{O}_{F_\mathfrak {p}}^\times \) (the ring of integers of the completion of \(F\) at \(\lambda _p\)), then the ideal \((\lambda _p)\) is unramified in \(K/F\) where \(K=F(\sqrt[p]{\xi })\).
Let \(K\) be a number field and \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \) for some \(\alpha \in K\). Then
where \(\sigma _i\) are the embeddings of \(K \) into \(\mathbb {C}\). Here \(\Delta (B)\) denotes the discriminant.
Let \(K\) be a number field with basis \(B=\{ b_1,\dots ,b_n\} \) and let \(\sigma _1,\dots ,\sigma _n\) be the embeddings of \(K\) into \(\mathbb {C}\). Now let \(M\) be the matrix
Then
Let \(R\) be a Dedekind domain, \(p\) a prime and \(\mathfrak {a},\mathfrak {b},\mathfrak {c}\) ideals such that
and suppose \(\mathfrak {a},\mathfrak {b}\) are coprime. Then there exist ideals \(\mathfrak {e},\mathfrak {d}\) such that
Let \(p \geq 3\) be an prime number, \(\zeta _p\) a \(p\)-th root of unity and \(K=\mathbb {Q}(\zeta _p)\). Assume that we have \(x,y,z \in \mathbb {Z}\) with \(\gcd (xyz,p)=1\) and such that
This is the so called "case I". To prove Fermat’s last theorem, we may assume that:
\(p \geq 5\);
\(x,y,z\) are pairwise coprime;
\(x \not\equiv y \mod p\).
Let \(K=\mathbb {Q}(\alpha )\) be a number field with \(n=[K:\mathbb {Q}]\) and let \(B=\{ 1,\alpha ,\alpha ^2,\dots ,\alpha ^{n-1}\} \). Then
where \(m_\alpha '\) is the derivative of \(m_\alpha (x)\) (which we recall denotes the minimal polynomial of \(\alpha \)).
Let \(p\) be a regular odd prime, \(\epsilon \in \mathbb {Z}[\zeta _p]^\times \) and \(n \in \mathbb {Z}_{\geq 1}\). Then the equation \(x^p+y^p+\epsilon (1-\zeta _p)^{pn}z^p=0\) has no solutions with \(x,y,z \in \mathbb {Z}[\zeta _p]\), all non-zero and \(xyz\) coprime to \((1-\zeta _p)\).