Basics of Field Extensions .... .... Ireland and Rosen, Ch 12

[Math Amateur]

I am reading Kenneth Ireland and Michael Rosen's book, "A Classical Introduction to Modern Number Theory" ... ...

I am currently focused on Chapter 12: Algebraic Number Theory ... ... I need some help in order to follow a basic result in Section 1: Algebraic Preliminaries ...

The start of Section 1 reads as follows:

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QUESTION 1


In the above text by Ireland and Rosen, we read the following:"... ... Suppose \(\displaystyle \alpha_1, \alpha_2, \ ... \ ... \ , \alpha_n\) is a basis for \(\displaystyle L/K\) and \(\displaystyle \alpha \in L\).Then \(\displaystyle \alpha \alpha_i = \sum_j a_{ ij } \alpha_j\) with \(\displaystyle a_{ ij } \in K\) ... ... "

My question is ... ... how do Ireland and Rosen get \(\displaystyle \alpha \alpha_i = \sum_j a_{ ij } \alpha_j\) ... ... ?

My thoughts are as follows ...Given \(\displaystyle L/K\), we have that \(\displaystyle L\) is a vector space over \(\displaystyle K\).

... we then let \(\displaystyle \alpha_1, \alpha_2, \ ... \ ... \ , \alpha_n\) be a basis for \(\displaystyle L\) as a vector space over \(\displaystyle K\)

( I take it that that is what I&R mean by "... ... Suppose \(\displaystyle \alpha_1, \alpha_2, \ ... \ ... \ , \alpha_n \)is a basis for \(\displaystyle L/K\)")... we then let \(\displaystyle \alpha \in L\) ... ... then there exist \(\displaystyle a_1, a_2, \ ... \ ... \ , a_n \in K\) such that \(\displaystyle \alpha = a_1 \alpha_1 + a_2 \alpha_2 + \ ... \ ... \ a_n \alpha_n\) so that \(\displaystyle \alpha \alpha_i = ( a_1 \alpha_1 + a_2 \alpha_2 + \ ... \ ... \ a_n \alpha_n ) \alpha_i\) ... ... ... (1)... BUT ...

Ireland and Rosen write (see above)\(\displaystyle \alpha \alpha_i = \sum_j a_{ ij } \alpha_j\)\(\displaystyle = a_{ i1 } \alpha_1 + a_{ i2 } \alpha_2 + \ ... \ ... \ + a_{ in } \alpha_n\) ... ... ... (2)My question is ... how do we get expression (1) equal to (2) ... ...
QUESTION 2In the above text by Ireland and Rosen, we read the following:"... ...The norm of \(\displaystyle \alpha, N_{ L/K } ( \alpha )\) is \(\displaystyle text{ det} (a_{ ij } )\) ... ... I cannot fully understand the process involved in forming the norm ... can someone please explain ... preferably via a simple example ...
Hope someone can help ... Peter

 

[Euge]

Hi Peter,

It appears you've missed something key here, which is that $L$ is a field. So all the products $\alpha \alpha_i$ belong to $L$. As $L$ is spanned by the $\alpha_j$, each product $\alpha \alpha_i$ is $K$-linear combination of the $\alpha_j$, which is displayed in the formulae

$$\alpha\alpha_i = \sum_{j} a_{ij} \alpha_j, \quad a_{ij}\in K$$

 

[Euge]

To answer your second question, consider $$L = \Bbb Q(\sqrt{2}) = \{a + b\sqrt{2} : a,b\in \Bbb Q\}\quad \text{and}\quad K = \Bbb Q$$ Set $\alpha = 3 + 2\sqrt{2}$. The numbers $1, \sqrt{2}$ form a basis for $L/K$. Further,

$\alpha \cdot 1 = 3 + 2\sqrt{2}$
$\alpha \cdot \sqrt{2} = 4 + 3\sqrt{2}$

Hence, the matrix $(a_{ij})$ of $\alpha$ with respect to the ordered basis $\{1,\sqrt{2}\}$ is given by

$$\begin{pmatrix}3 & 2\\ 4 & 3\end{pmatrix}$$

The determinant of this matrix is $3(3) - 4(2) = 1$, so the norm of $\alpha$, $N_{L/K}(\alpha)$, equals $1$.

As an exercise, show that in general, the norm of an element $a + b\sqrt{2}$ in $Q(\sqrt{2})$ is $a^2 - 2b^2$.

 

[Math Amateur]

Hi Peter,

It appears you've missed something key here, which is that $L$ is a field. So all the products $\alpha \alpha_i$ belong to $L$. As $L$ is spanned by the $\alpha_j$, each product $\alpha \alpha_i$ is $K$-linear combination of the $\alpha_j$, which is displayed in the formulae

$$\alpha\alpha_i = \sum_{j} a_{ij} \alpha_j, \quad a_{ij}\in K$$

Thanks Euge ... yes ... should have known!

Obvious now, of course ... :(

Peter

 

[Math Amateur]

To answer your second question, consider $$L = \Bbb Q(\sqrt{2}) = \{a + b\sqrt{2} : a,b\in \Bbb Q\}\quad \text{and}\quad K = \Bbb Q$$ Set $\alpha = 3 + 2\sqrt{2}$. The numbers $1, \sqrt{2}$ form a basis for $L/K$. Further,

$\alpha \cdot 1 = 3 + 2\sqrt{2}$
$\alpha \cdot \sqrt{2} = 4 + 3\sqrt{2}$

Hence, the matrix $(a_{ij})$ of $\alpha$ with respect to the ordered basis $\{1,\sqrt{2}\}$ is given by

$$\begin{pmatrix}3 & 2\\ 4 & 3\end{pmatrix}$$

The determinant of this matrix is $3(3) - 4(2) = 1$, so the norm of $\alpha$, $N_{L/K}(\alpha)$, equals $1$.

As an exercise, show that in general, the norm of an element $a + b\sqrt{2}$ in $Q(\sqrt{2})$ is $a^2 - 2b^2$.

Thanks Euge ... wonderfully clear and helpful example ...

Peter