# I Field Extensions and "Free Parameters"

1. May 30, 2017

### Math Amateur

I am reading "Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with some remarks of Lovett pertaining to Theorem 7.1.10 ...

The statement of Theorem 7.1.10 reads as follows (page 325) :

The remarks pertaining to Theorem 7.1.10 read as follows (page 326) :

I do not understand the use of 't' in the above text ... nor do I fully understand the analysis involving it ...

My specific question is as follows:

In the above remarks, Lovett writes the following:

" ... ... For example keeping $t$ as a free parameter, $F[t]$ is a subring of $F(t)$. ... ... "

What is $'t'$ and why exactly are we introducing it?

Why not stick with $x$ and $F[x]$ and $F(x)$ ... ... ?

I note that Lovett does not usually introduce a "free parameter" (whatever that is?) and happily deals with the indeterminate $x$... ... ?? ... ... so ... indeed, one may ask when is a "free parameter" necessary and when is it not needed ...

Hope someone can help ...

Peter

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2. May 31, 2017

### andrewkirk

$F(\alpha)$ is the field generated by $F\cup\{\alpha\}$, and $F[x]$ is the ring of polynomials with coefficients in $F$.

But what does $F[\alpha]$ denote? A quick internet search for use of that notation draws a blank.

Does the author explain what he means by $F[\alpha]$? I feel we need to make sense of that before we try to make sense of $F[t]$.

Failing that, does the other book you were reading about this recently (Dummitt & Foote?) use that notation and if so do they define it? I have a vague memory of seeing something like $F[\alpha]$ in one of your earlier posts about the other book, but I may be misremembering.

3. May 31, 2017

### Staff: Mentor

In general $\mathbb{F}[.]$ denotes a ring over $\mathbb{F}$ and $\mathbb{F}(.)$ denotes a field, the quotient field of $\mathbb{F}[.]$.

In case it is $\mathbb{F}[t]$ or $\mathbb{F}[x]$ we usually mean indeterminates, i.e. a variable $t$ or $x$ and polynomial rings where $t^{-1} \notin \mathbb{F}[t]$ and $x^{-1} \notin \mathbb{F}[x]$, and their quotient fields where $t^{-1} \in \mathbb{F}(t)$, resp. $x^{-1} \in \mathbb{F}(x)$.

The indeterminates behave like transcendental numbers, as they don't obey any algebraic equation. The only difference is an intuitive one: we do not replace numbers - algebraic or transcendental - by other numbers, but we are used to replace indeterminates by numbers, such as $\mathbb{F}[x] \longrightarrow \mathbb{F}[\sqrt{2}]$.

Now what the author means is, that $\mathbb{Q}[\pi]$ and $\mathbb{Q}(\pi)$ are not the same algebraic object. The first is only a ring, an infinte ring extension of the rationals, because $\frac{1}{\pi} \notin \mathbb{Q}[\pi]$, whereas the second is a field: $\frac{1}{\pi} \in \mathbb{Q}(\pi)$.

On the other hand, the situation changes, if we talk about algebraic numbers. In this case is $\mathbb{F}[\alpha] = \mathbb{F}(\alpha)$.
Can you show, why $\frac{1}{\alpha} \in \mathbb{F}[\alpha]$ if $\alpha$ is algebraic over $\mathbb{F}$?

4. May 31, 2017

### andrewkirk

I'm afraid I still don't follow this. Are you able to write a formal definition of the symbol string $F[\alpha]$?

A formal definition of the symbol string $F[x]$, where $F$ denotes a field, is as the set $F^*\triangleq \bigcup_{k=1}^\infty F^k$ (ie the set of all finite $k$-tuples of elements of $F$, where $k$ is any positive integer) with operations of addition and multiplication defined so as to correspond with those operations for polynomials that have coefficients equal to the components of the $k$-tuples.

With this definition the symbol string $[x]$ is a postfix, unary operator that maps a field to its corresponding polynomial ring. Writing any letter other than $x$ between square brackets is an undefined symbol string. So we need a new definition to make sense of symbol strings like $F[\alpha]$ or $F[t]$.

Thank you

5. May 31, 2017

### Staff: Mentor

Another remark: Sometimes the quotient fields $\mathbb{F}(x)$ or $\mathbb{Q}(\pi)$ are written $\mathbb{F}[x,\frac{1}{x}]$ and $\mathbb{Q}[\pi,\frac{1}{\pi}]$ which explicitly makes sure that the inverse elements are adjoint as well.

6. May 31, 2017

### Staff: Mentor

Yes, it's the image of the projection $F[x] \twoheadrightarrow F[x]/(m_\alpha(x))$ where $m_\alpha(x)$ is a unitary irreducible polynomial over $F$. The elements in this image can be written as polynomials in $\alpha$ where a certain polynomial equation equals zero, namely $m_\alpha(x)=0$ in which case we call the zeros of it, e.g. $\alpha$ algebraic. All "numbers" $\tau$ which do not have such an equation satisfy $F[x] \cong F[\tau]$ and are called transcendental numbers. Numbers in this context refer to elements in some field extension of $F$ which has to be given otherwise. One could also start with such a field extension, i.e. a given algebraic number $\alpha$ and define $F[\alpha]$ as the image of $F[x] \longrightarrow F[\alpha]$ by substitution with the arithmetic in the extension field.
The most formal definition of a polynomial ring, yes.
$x$ is not always the indeterminant. Some authors use $t$. I don't think that it is an undefined symbol string, as for writing it $x, t$ or $\alpha$ is only a convention. So formally $F[\alpha]$ would be the polynomial ring in $\alpha$ over $F$. But I agree, that $\alpha$ usually denotes a number in some extension field of $F$ that satisfies a polynomial equation, i.e. the quotient ring. So formally we introduce an algebraic dependency among the coefficients of elements in $F[x]$.

I'm not sure if this answers the question. I only wanted to say that (usually) $F[.]$ brackets denote rings whereas $F(.)$ parantheses denote their quotient fields. In case of algebraic numbers they are the same, in case of transcendental numbers (or indeterminants) they are not.

7. May 31, 2017

### Math Amateur

Thanks Andrew and fresh_42 ... the picture is becoming clearer ...

I just thought I would present Lovett's explanation of $F[ \alpha ]$ and $F( \alpha )$ ... and also $F[x]$ ...

Lovett's first significant mention of such symbols is in Section 5.2: Rings Generated by Elements (pages 216 - 225) ...

In Section 5.2.1 Generated Subrings, Lovett defines $R$ where $R$ is a subring of a commutative ring $A$, and $S$ is a subset of $A$ ... ... as follows:

Lovett then goes on to define polynomial rings $R[x]$ ... in section 5.2.2 ...

The start of Section 5.2.2 reads as follows:

Lovett then goes on to define/discuss addition and multiplication of polynomials ...

I note that in defining polynomial rings Lovett uses the term "the variable $x$" and not the mysterious term "indeterminate" for $x$ ... ...

In the context of fields, Lovett mentions $F[ \alpha ]$ and $F( \alpha )$ in Section 7.1 Introduction to Field Extensions ... . The reference is in the context of ways to construct some field extensions ... and reads as follows:

Hope the above helps ...

Thanks again for your clarifications and help ...

Peter

EDIT ...I have no idea why my text is crossed out ???

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• ###### Lovett - F[a] and F(a) - Constructing Field Extensions ....png
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8. May 31, 2017

### Math Amateur

Hi Andrew, fresh_42 ... ...

In order to further clarify my puzzlement at the differences between $F[x]$ and $F[t]$ I thought i would present the theorem which first convinced me I need a better understanding of t as a "free parameter" ... ...

The Theorem concerned is Theorem 7.6.3 in Lovett, Section Splitting Fields and Algebraic Closure ... ... the theorem and its proof read as follows:

In the text above from Lovett, in the third paragraph of the proof, we read the following:

" ... ... In $F'[t], (t - \overline{x} )$ is a linear factor of $f(t)$ ... ... "

I am not sure why we are introducing $t$ ... and, further, I am not sure of the nature of $t$ ...

Is it just a variable in $F'$ where we want to distinguish it from $x$ which is a variable in $F$ ... is that all there is to it ..

Mind you, later we read ...

" ... ... If

$f(x) = \sum_{ i = 0 }^{ n + 1 } a_i x^i$

then $f(t) = 0$ is equivalent to $f(t) - f( \overline{x} ) = 0$ ... ... "

I cannot see why $f(t) = 0$ is equivalent to $f(t) - f( \overline{x} ) = 0$ ... ... ?

Can you help?

Peter

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• ###### Lovett - Theorem 7.6.3 and proof .....png
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9. May 31, 2017

### andrewkirk

OK it's starting to take shape now. This is how it looks to me, based on yours and fresh's posts:

• Let $F$ be a field that is a sub-field of field $E$.