Math 361, Spring 2016, Assignment 7

Carefully define the following terms, then give one example and one non-example of each:

1. Finitely generated (field extension).
2. Vector space (over a field $F$).

Solve the following problems:

1. Consider the field $\mathbb{Q}(\sqrt{2},\pi)$, the smallest subfield of $\mathbb{R}$ containing $\sqrt{2}$ and $\pi$. Describe the elements of this field, then give sample calculations in which you add and multiply elements of the field.
2. (Lines) In classical linear algebra, the set of all scalar multiples of a given non-zero vector $v$ is called the line spanned by $v$. Now consider the field $\mathbb{Z}_3[x]/\left\langle x^2+1\right\rangle$, regarded as a vector space over $\mathbb{Z}_3$. As usual, let $\alpha$ denote the coset $x+\left\langle x^2+1\right\rangle$. List the elements of the line spanned by $1+2\alpha$.
3. (Gauss-Jordan elimination) In classical linear algebra, the Gauss-Jordan algorithm uses elementary row operations to find the "reduced row-echelon form" of a matrix $M$. If necessary, review this algorithm in a linear algebra text, and convince yourself that it does not really depend on the entries of the matrix coming from $\mathbb{R}$; in fact the algorithm can be executed if the entries come from an arbitrary field $F$. For example, taking $F=\mathbb{Z}_3$, find the reduced row-echelon form of the matrix $$\begin{bmatrix}2 & 1 & 0 \\ 2 & 0 & 1\end{bmatrix}$$.
4. Re-read your linear algebra textbook, identifying all the problems and examples that use Gauss-Jordan elimination to calculate something. Which of these would work with coefficients coming from an arbitrary field? Is there anything in an elementary linear algebra class that doesn't work over an arbitrary field? (Hint: yes.)

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Solutions:

Vocab

1. A field extension $(F, E, \iota)$ is finitely generated if $E=F(S)$ (i.e. $E$ is the subextension of $F$ generated by $S\in F$) for some

finite set $S$.

• e.g. $\mathbb{Q}(\{\sqrt{2}\})$ is finitely generated, as the set containing $\sqrt{2}$ has cardinality $1$, which is finite
• $\neg$e.g. $\mathbb{Q}(\mathbb{R})$ is not finitely generated, because $\mathbb{R}$ has infinite cardinality

Not just infinite, but actually uncountable. (Using tedious arguments from set theory one can show that any simple extension of a countable field must be countable, and hence by induction any finitely generated extension of a countable field must be countable. Thus $\mathbb{R}$ cannot be finitely generated over $\mathbb{Q}$. On the other hand, there are plenty of merely infinite fields that are finitely generated over $\mathbb{Q}$, for example $\mathbb{Q}$ itself.) -Steven.Jackson (talk) 15:38, 25 March 2016 (EDT)

2. A vector space over a field $F$ is a set of vectors $V$, together with operation $+$, element $0$, and map $\mu :F\times V\rightarrow V$

such that...

a. $(V, +, 0)$ is an abelian group

b. $\mu$ is the "scalar multiplication map," with properties

i. $\mu(k, v)$ is usually abbreviated $k\cdot v$ or $kv$

ii. $k(v_1 + v_2)=kv_1+kv_2$ and $(k_1 + k_2)v = k_1v + k_2v$

iii. $0\cdot v=0$ and $1_F\cdot v=v$

• e.g. Let $F$ be an arbitrary field with extension $E=V$ by $\iota$. This is a vector space with $\mu(k,v):k\cdot v\mapsto \iota(k)\cdot v$
• $\neg$e.g. Let $F=\mathbb{R}$, and $V=$GL$_2(\mathbb{R})$. This is not a vector space because GL$_2(\mathbb{R})$ is not abelian