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:

View attachment 6509
QUESTION 1


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

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

My thoughts are as follows ...Given $$L/K$$, we have that $$L$$ is a vector space over $$K$$.

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

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

Ireland and Rosen write (see above)$$\alpha \alpha_i = \sum_j a_{ ij } \alpha_j$$$$= a_{ i1 } \alpha_1 + a_{ i2 } \alpha_2 + \ ... \ ... \ + a_{ in } \alpha_n$$ ... ... ... (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 $$\alpha, N_{ L/K } ( \alpha )$$ is $$text{ det} (a_{ ij } )$$ ... ... 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