# I Basic Notation for Field Extensions ...

1. Apr 10, 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 the meaning of some of the basic notation ...

In talking about ways to construct field extensions Lovett writes the following on page 322, Chapter 7 ... ...

In the above text from Lovett, he writes:

"... ... If $\alpha \in R - F$, then $F[ \alpha ]$ is the subring of $R$ generated by $R$ and $\alpha$.** See Section 5.2.1. Since $F[ \alpha ]$ is an integral domain, we can take the field of fractions $F( \alpha )$ of $F[ \alpha ]$. See Section 6.2. ... ...

(** See note at end of post ... )

"My problem is that the way Lovett seems to define $F[ \alpha ]$ and $F[ \alpha ]$ seems (on the surface, at least) to be different from all the other texts I am reading ... for example Dummit and Foote or Gallian ...

Lovett's definition of the notation $F[ \alpha ]$ comes in sections 5.2.1 and 5.2.2 where he defines rings generated by elements as follows ... ...

He goes on from the above in the next section to define rings of polynomials using the same notation as he did for generated subrings ... as follows:

Lovett defines $R(x)$ in Section 6.2.2 as the field of fractions of $R[x]$ ... as follows:

Dummit and Foote, on the other hand (like a number of other algebra texts) define $F( \alpha )$ as follows:

Further, Dummit and Foote simply define $R[x]$ in terms of polynomial rings ... as follows:

Can someone please give an explanation of the apparently different definitions between Lovett and the other texts on this subject ...

[Please excuse me moving between rings and fields in the above definitions ... ]

Hope someone can help ...

Peter

================================================================================

NOTE

** Where Lovett writes

" ... ... If $\alpha \in R - F$, then $F[ \alpha ]$ is the subring of $R$ generated by $R$ and $\alpha$."

Is this correct?

" ... ... If $\alpha \in R - F$, then $F[ \alpha ]$ is the subring of $R$ generated by $F$ and $\alpha$."

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• ###### D&F - Defn of R[X] as ring of polynomials ....png
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Last edited: Apr 11, 2017
2. Apr 11, 2017

### Staff: Mentor

No.
Yes. It's a typo.

In general variables or better indeterminants are noted by the letters $x,y,z$ or $t$ or these letters in caps. They behave like a transcendental extension as they do not satisfy any algebraic equation. However, there is no "law" that this is always the case, so in the end it depends on the author.

In contrast are algebraic numbers, i.e. those which satisfy an algebraic equation, like $i$ satisfies $x^2+1=0$ or $\sqrt{2}$ satisfies $x^2-2=0$. Those algebraic numbers are usually denoted by small Latin or Greek letters from the beginning of the alphabet.

If $F \subseteq R$ are rings, say integral domains, and $\alpha \in R-F$, then:

$F[x] =$ polynomial ring with coefficients in $F$ and indeterminant $x$ = transcendental simple ring extension of $F$

a) If there is no algebraic equation for $\alpha$ then $F[\alpha] =$ transcendental simple ring extension of $F$ with transcendental element $\alpha \;;\; F[\alpha] \cong F[x]$. An example is $\mathbb{Q}[\pi] \cong \mathbb{Q}[x]$.

b) If there is a polynomial $f(x) \in F[x]$ with $f(\alpha)=0$, then $\alpha$ is algebraic over $F$. In this case
$F[\alpha] =$ algebraic simple ring extension of $F$ with algebraic element $\alpha$. If we chose $f(x)$ irreducible then $F[\alpha]\cong F[x]/<f(x)>$.

The quotient fields (those which are what $\mathbb{Q}$ is to $\mathbb{Z}$) are denoted by round brackets:
$F(x) =$ quotient field of the polynomial ring $F[x]$, i.e. quotients of polynomials in $x$ with coefficients in $F$
$F(\alpha)$ is the quotient field of $F[\alpha]$.
A special notation for the quotient field of $F$ or $R$ alone doesn't exist.

Furthermore there are the following similar notations:
$F[[x]] =$ formal power series with coefficients in $F$, i.e. "endless polynomials"
$F((x)) =$ formal Laurent series with coefficients in $F$, i.e. the quotient field of the ring $F[[x]]$
(cp. https://en.wikipedia.org/wiki/Formal_power_series)

3. Apr 11, 2017

### Math Amateur

Thanks fresh_42 ... really appreciate your help on this matter ...

Still reflecting on what you have written ...

Peter

4. Apr 11, 2017

### Math Amateur

Hi fresh_42,

Thanks again for your post ... it cleared up a number of points ...

But just trying to completely understand how the definitions Lovett and Dummit and Foote of $F( \alpha )$ and $F[ \alpha ]$ are reconciled ...

Dummit and Foote

Adapting D&F's definition a bit ... we have that if $K$ is an extension field of $F$ and an element $\alpha \in K$ then ... according to D&F ... :

$F( \alpha )$ is the smallest subfield of $K$ containing both $F$ and the element $\alpha$ ... that is $F( \alpha )$ is the field generated by $F$ and $\alpha$ ... ...

... ... while ... ...

Lovett

Lovett says that if $R$ is an integral domain (or, I guess, a field) containing a field $F$ then ...

$F[ \alpha ]$ is the subring (subfield) of $R$ generated by $F$ and $\alpha$ ....

So ... my question is as follows:

How do we explain how it is that

$F( \alpha )$ and $F[ \alpha ]$ seem to be defined (by two different texts) as the same thing ... ...

Can you help?

Peter

5. Apr 11, 2017

### Staff: Mentor

This is correct. We need the least all elements of $F$, all powers of $\alpha$ and all quotients of them. On the other hand this defines a field $F\subseteq F(\alpha) \subseteq K$, so it's the smallest with these properties.
You must not assume a field. Maybe $F$ is a field in the ring $R$, as $\mathbb{Q} \subseteq \mathbb{Q}[\pi]$ is a subfield in a ring, too. But $F[ \alpha ]$ only contains the polynomials in $\alpha$, which means you don't have the quotients with $\alpha$ automatically included: $\alpha^n \in F[ \alpha ] \, , \,n \in \mathbb{N}\, , \,$ but $\frac{1}{\alpha} \notin F[\alpha ]$. At least not in general like $\frac{1}{\pi} \notin \mathbb{Q}[\pi]$. However this example heavily depends on the fact, that $\pi$ and $\frac{1}{\pi}$ are transcendental numbers and you cannot express one by the other with polynomials; you need them both to get a field.

Things change, if $\alpha$ is algebraic over $F$, that is it satisfies an equation $0=a_n\alpha^n + a_{n-1}\alpha^{n-1} + \ldots + a_1\alpha+a_0$ with coefficients $a_k$ in the field $F$.

I leave it as an exercise for you to show that $F[\alpha] = F(\alpha)$ in this case.

6. Apr 12, 2017

### Infrared

Nice post! I just thought I'd comment that $\mathbb{Q}[\pi,\frac{1}{\pi}]$ still isn't a field.

7. Apr 12, 2017

### Staff: Mentor

Give me a hint. Why not?

8. Apr 12, 2017

### Infrared

$\pi+1$ doesn't have an inverse.

9. Apr 12, 2017

### Math Amateur

Thanks fresh_42 ... that post was particularly helpful ...

Your post was particularly helpful for the remarks regarding $F[\alpha]$ ...

In this matter you write:

"... ... You must not assume a field. Maybe $F$ is a field in the ring $R$ ... ... "

... but ... I note that Lovett insists on $R$ being an integral domain, at least when we are in the context of field extensions ...

Can you comment?

Peter

10. Apr 13, 2017

### Staff: Mentor

Hi Peter,

it isn't important what $R$ is, only that it provides an element $\alpha \notin F$, so ring is sufficient for $R$. But integral domain is convenient, because otherwise one would have to deal with special cases such as whether $\alpha$ is a zero divisor or not.

For $F[\alpha] = F(\alpha)$ (with $\alpha$ algebraic) on the other hand, you need $F$ to be a field, or at least that the minimal polynomial of $\alpha$ behaves nicely because in order to calculate $\frac{1}{\alpha} \in F[\alpha]$ we need divisions in $F$.

I assume Lovett is steering towards field extensions. Of course one can always stay in the world of rings and consider the situation $\alpha \notin F \subseteq R \ni \alpha$, i.e. ring extensions. $F[\alpha]$ is then a ring in between: $F \subseteq F[\alpha] \subseteq R$ and we would consider them as modules, resp. eventually ideals which are the appropriate tools in ring theory.

To summarize: $F$ a field and $R$ an integral domain (which allows a quotient field!) are somehow minimal conditions to proceed toward field extensions in a way, that doesn't split into a vast number of special cases. Otherwise we would do module and ideal theory and fields would be nothing else than the factor ring of a maximal ideal.

11. Apr 13, 2017

### Math Amateur

Hi fresh_42,

Thanks for the help ...

BUT ... just a clarification ...

You write:

" ... ... it isn't important what $R$ is, only that it provides an element $\alpha \notin F$ ... ... "

However it is important that $R$ is not a field ... that is correct isn't it ... ... (at least in defining/constructing $F[\alpha]$, anyway)

Peter

12. Apr 13, 2017

### Staff: Mentor

Why do you think so? It would rule out all field extensions, the entire Galois theory and most of all: what for?

We want to consider situations like $\mathbb{Q} \subseteq \mathbb{Q}[\sqrt[3]{2}] \subseteq \mathbb{R}$ and many more, where all involved sets are fields. We need $R$ to be a ring, because $F[\alpha]$ are polynomials, so we want to multiply and add, but fields are also rings.

$R$ being an integral domain, but not a field, would mean there is an ideal $P$ in a ring $S$, which contains $R$, such that $R \cong S/P$ and $P \subseteq S$ is a prime ideal but not a maximal ideal. What would this be good for, except some funny exercises?

13. Apr 14, 2017

### Math Amateur

Hi fresh_42,

When I was talking about $R \subset F$ where R is a ring or, better, an integral domain, and $F$ is a field ... ... I was talking only in the context of the construction of $F [ \alpha ]$, the ring (or integral domain) of polynomials ... to construct $F [ \alpha ]$ we construct the subring of $R$ generated by $F$ and $\alpha$ ... we do not, in this first stage want $R$ to be a field, or as you have said we will have elements like $1/ \alpha$ in $F [ \alpha ]$ which we do not want ...

BUT ... I must say that I did not make this clear ... apologies

... after we have constructed $F [ \alpha ]$, we can then take the field of fractions of $F [ \alpha ]$ and attain $F ( \alpha )$ ... ...

BUT ... I am aware that in the theory of field extensions it is often the case that we consider fields $K$ and $F$ where $F \subset K$ ...

Hopefully the above reasoning makes sense and is not confused or confusing ...

Peter