2.2.23 Discriminants of trinomials
Let \(\alpha \) be an algebraic number with minimal polynomial \(m_\alpha \) and let \(\beta \in \mathbb {Q}(\alpha )\) be such that \(\alpha =\frac{a}{b\beta +c}\) with \(a,b,c \in \mathbb {Q}\) and \(b \neq 0\). Then \(\deg (m_\alpha )=\deg (m_\beta )\).
Note that \(\mathbb {Q}(\alpha )=\mathbb {Q}(\beta )\) (just do double inclusion). Then since \(\deg (m_\beta )=[\mathbb {Q}(\beta ):\mathbb {Q}]=[\mathbb {Q}(\alpha ):\mathbb {Q}]=\deg (m_\alpha )\) (by Cor. 1.3.12) we get the result.
Let \(K=\mathbb {Q}(\alpha )\) a number field with \(m_\alpha (x)=x^n+ax+b\). Then
From Proposition 2.2.19 we have
Now, let \(\beta =m_\alpha '(\alpha )\), then we have
Since \(\alpha ^n+a\alpha +b=0\) implies \(n\alpha ^{n-1}=-na-nb\alpha ^{-1}\). Which gives
Now, using Lemma 2.2.24 we see that \(n=\deg (m_\alpha )=\deg (m_\beta )\). Moreover, if we take \(m_\alpha (x)=x^n+ax+b\), evaluate at \(\frac{-nb}{\beta +(n-1)a}\) and clear denominators, we get that
Therefore \(\beta \) is a root of
which is a monic polynomial of degree \(n=\deg (m_\beta )\), therefore this is the minimal polynomial of \(\beta \).
Now, by Proposition 1.6.6 or Corollary 1.6.8 we have \(C_\beta =m_\beta \). Therefore by Proposition 1.7.3 we see that \((-1)^n\) times the constant coefficient of \(m_\beta \) is \(N_{K/\mathbb {Q}}(\beta )\). Therefore we get
which gives the result.
Let \(K=\mathbb {Q}(\alpha )\) be a number field with \(m_\alpha (x)=x^3+ax+b\). Then
Let \(K=\mathbb {Q}(\alpha )\) a number field with \([K:\mathbb {Q}]=3\). Then \(m_\alpha (x)=x^3+ax^2+bx+c\) for \(a,b,c \in \mathbb {Q}\). If we replace \(\alpha \) with \(\alpha ':=\alpha +a/3\) then \(\mathbb {Q}(\alpha )=\mathbb {Q}(\alpha ')\) but this has the effect of removing the \(x^2\) term in the minimal polynomial, so after relabelling if necessary, we have \(K=\mathbb {Q}(\alpha )\) with \(m_{\alpha }(x)=x^3+ax+b\). Therefore Corollary 2.2.26 gives us a quick way to find the discriminant of a cubic field.
Let \(\alpha \) be a root of \(x^3+x+1\), and let \(K=\mathbb {Q}(\alpha )\). Then \(\Delta (\{ 1,\alpha ,\alpha ^2\} )=-31\) which is square-free and therefore by Corollary 2.2.9 \(\{ 1,\alpha ,\alpha ^2\} \) is an integral basis and therefore \(\mathcal{O}_K=\mathbb {Z}[\alpha ]\).