Field extension basis multiplication in 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:

QUESTION 1In 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

 

[andrewkirk]

In Question 1, we observe that, since $\alpha_1,...,\alpha_n$ form a basis for $L$ as a vector space over $K$, any element of $L$ can be written as a linear combination of those basis elements, with coefficients in $K$, that is, as $\sum_{j=1}^na_j\alpha_j$ with $a_j\in K\forall j$.

Since $\alpha,\alpha_i$ are both in $L$, which is a field, $\alpha\alpha_i$ must also be in $L$ and hence can be written as such a linear sum. We then just relabel each $a_j$ as $a_{ij}$ and we have the text's formula.

In question 2, note that, given $\alpha\in L$ and a basis $\alpha_1,...,\alpha_n$ for $L$, each $\alpha_i$ gives us a set of $n$ coefficients in $K$: $a_{i1},...,a_{in}$. Since there are $n$ $\alpha_i$s, we can put those coefficients in a $n\times n$ matrix and then calculate a determinant of that matrix.

 

[Math Amateur]

Thanks Andrew ... just reflecting on what you have written ...

Peter