1.6 The standard representation
Let \(K\) be a number field and let \(\alpha \in K\). Then we can associate to \(\alpha \) a linear operator
This can be written as a matrix over \(\mathbb {Q}\) by picking a basis of \(K\) over \(\mathbb {Q}\). The map \(\alpha \to A_\alpha \) is called the standard representation.
If we had \(K/F\) an extension of number fields then by picking a basis of \(K/F\) we can write \(A_\alpha \) as a matrix with coefficients in \(F\).
Let \(K=\mathbb {Q}(\sqrt{5})\) and let \(\alpha =3+2\sqrt{5}\). We have a basis for \(K/\mathbb {Q}\) given by \(\{ 1,\sqrt{5}\} .\) So \(A_\alpha (1)=3+2\sqrt{5}\) and
therefore in this basis, we have
Let \(K\) be a number field and \(\alpha \in K\).
The map \(\alpha \to A_\alpha \) is injective.
Let \(\alpha ,\beta \in K\) and \(\lambda \in \mathbb {Q}\) then
\[ A_{\alpha \beta }=A_{\alpha }A_{\beta } \qquad A_{\alpha +\beta }=A_\alpha + A_\beta \qquad A_{\lambda \alpha }=\lambda A_\alpha \]If \(C_\alpha \) is the characteristic polynomial of \(A_\alpha \) then \(C_\alpha (\alpha )=0\).
For [1] we just need to note that \(A_\alpha (1)=\alpha \), therefore \(\alpha =\beta \) if and only if \(A_\alpha =A_\beta \).
For [2], this is obvious from the definition of \(A_\alpha \) as the multiplication by \(\alpha \) map.
Lastly for [3] note that by Cayley–Hamilton that \(C(A_\alpha )=0\). Now, from [2] it follows that for any polynomial \(f(x) \in \mathbb {Q}[x]\) we have \(f(A_\alpha )=A_{f(\alpha )}\), so \(A_{C_\alpha (\alpha )}=C_\alpha (A_\alpha )=0\) but then by [1] we must have \(C_\alpha (\alpha )=0\).
For \(K\) a number field and \(\alpha \in K\), then the characteristic polynomial of \(A_\alpha \) is known as the field polynomial of \(\alpha \). We shall denote it by \(C_\alpha \), but note that it depends on the field we are working with. In particular, if we write a basis for \(K\) over \(\mathbb {Q}\) or a basis for \(K\) over \(F\) (for \(F\) some subfield), then the field polynomial can be different in each case. Therefore, unless otherwise stated we will assume we mean a basis of \(K/\mathbb {Q}\).
Let \(K=\mathbb {Q}(\alpha )\) for some algebraic number \(\alpha \). Then \(C_\alpha =m_\alpha \), in other words, the field polynomial of \(\alpha \) agrees with the minimal polynomial.
Note that \(C_\alpha \) will be a monic polynomial of degree \([K:\mathbb {Q}]\) and has \(\alpha \) as a root. But this means it is divisible by \(m_\alpha \). But the are both monic and have the same degree, therefore must be equal.
Let \(K\) be a number field and \(\beta \in K\). Then \(A_\beta \) is diagonalizable over \(\mathbb {C}\) and
Here the product is over all embeddings \(\sigma _i\) of \(K\) into \(\mathbb {C}\).
By the Primitive Element Theorem 1.5.11 we can write \(K \cong \mathbb {Q}(\alpha )\). Now, from Proposition 1.6.6, we know \(C_\alpha =m_\alpha \) and by Proposition 1.5.6 we know that \(m_\alpha =\prod _{i}(x-\sigma _i(\alpha ))\), therefore, \(C_\alpha \) has \(n:=[K:\mathbb {Q}]\) distinct roots (by Lemma 1.5.7), therefore \(A_\alpha \) is diagonalisable over \(\mathbb {C}\), with eigenvalues being the conjugates of \(\alpha \). So we can find a matrix \(P\) (over \(\mathbb {C}\)) such that
Now, \(\beta \in K\) we can find \(f(x) \in \mathbb {Q}[x]\) such that \(f(\alpha )=\beta \) and therefore
\begin{align*} P^{-1} A_\beta P=P^{-1} A_{f(\alpha )} P= P^{-1} f(A_\alpha ) P=f(P^{-1} A_\alpha P)& = \left(\begin{matrix} f(\sigma _1(\alpha )) & & \\ & \ddots & \\ & & f(\sigma _n(\alpha )) \end{matrix} \right)\\ & = \left(\begin{matrix} \sigma _1(\beta ) & & \\ & \ddots & \\ & & \sigma _n(\beta ) \end{matrix} \right) \end{align*}And since \(P^{-1}A_\beta P\) and \(A_\beta \) have the same characteristic polynomial, we get the result.
Let \(K\) be a number field and \(\beta \in K\), then
By Proposition 1.6.7 we know all the roots of \(C_\beta \) are of the form \(\sigma _i(\beta )\). Now, let \(v_i\) be the embedding of \(\mathbb {Q}(\beta )\). Then from Proposition 1.5.8 we know that each \(v_i\) extends to \([K:\mathbb {Q}(\beta )]\) embeddings of \(K\) and by definition each of these embeddings keeps \(v_i(\beta )\) the same. So for each \(v_i(\beta )\) there are \([K:\mathbb {Q}(\beta )]\) embeddings \(\sigma _j\) such that \(\sigma _j(\beta )=v_i(\beta )\). Now, since \(m_\beta (x)=\prod _i (x-v_i(\beta ))\) we see that
giving the result.
Note that \([K:\mathbb {Q}]=[K:\mathbb {Q}(\beta )][\mathbb {Q}(\beta ):\mathbb {Q}]\) so the power is correct, since \(C_\beta \) has degree \([K:\mathbb {Q}]\) and \(m_\beta \) has degree \([\mathbb {Q}(\beta ):\mathbb {Q}]\).