1. Anybody who has done a course in abstract algebra should be familiar with rings and fields. Let’s start with the definitions.

A ring is a set R together with binary operations + (addition) and × (multiplication) such that
1. R is an Abelian group under + with identity element 0 (called zero),
2. × is associative, and is distributive over + (both from the left and from the right)
Some authors also insist that a ring R must contain a multiplicative identity 1 (called unity).

A field is a set F with similar addition and multiplication operations such that
1. F is a ring (with unity) under + and ×,
2. F\{0} forms an Abelian group under ×.
Thus all fields are rings: {rings} ⊇ {fields}.

However the chain is much more interesting than that.

Note that ring multiplication need not be commutative. If it is, we say that the ring is a commutative ring. Matrices (i.e. square matrices) provide the best examples of noncommutative rings, while the integers furnish the most familiar example of a commutative ring.

Moreover, with matrices, it is possible for the product of two nonzero matrices to be zero. This is not the case with integers: if m and n are integers, and mn = 0, then either m = 0 or n = 0.

An integral domain is a commutative ring R (with unity) such that .

Another property of the integers is the unique factorization: every integer n > 1 can be written uniquely as a product of primes. This idea can be extended to integral domains. First, we need some definitions. (We’ll also drop the × symbol for ring multiplication and just write ab for a×b.)

If R is a commutative ring (with unity) and , we say that a divides b iff there exists such that . A unit (not to be confused with “unity”ť) in R is an element that divides 1. An element is said to be an irreducible iff r is not 0 or a unit and for any either a or b is a unit.

Now suppose given we can write where the ’s are irreducibles; suppose moreover that if , where the ’s are irreducibles, then m = n and there is a permutation of {1,…,m} such that for some unit . Then we say that a is uniquely factorized in R. A unique-factorization domain (or UFD) is an integral domain in which every element that is not 0 or a unit can be uniquely factorized.

Hence is a UFD. The units of are ±1, and the irreducibles are ±p, where p is a prime number. Actually, in ring theory, primes are defined differently from the way integer prime numbers are defined. If R is a commutative ring with unity, an element is a prime iff p is not 0 or a unit, and for any whenever p divides ab, it follows that p divides a or p divides b. It turns out that every prime is an irreducible in an integral domain; in a UFD, every irreducible is also a prime.

If R is a commutative ring, an ideal in R is a subset I of R such that I is an additive subgroup of R and . If there is an element such that , then I is said to be a principal ideal (generated by ). An integral domain in which every ideal is principal is called a principal-ideal domain (or PID).

is a PID. Given any integer n, the set is an ideal; conversely every ideal in is of this form. As a matter of fact, every PID is a UFD.

Yet another familiar property of the integers is the Euclidean property: if m and n are integers and n ≠ 0, then there exist integers q and r such that m = qn + r and 0 ≤ r < |n|.

In general, given a ring R, suppose there exists a function (called a norm function); if such that and either r = 0 or , then R is said to satisfy the Euclidean property. (For the integers, the norm function is the absolute-value function.) A Euclidean domain is an integral domain that satisfies the Euclidean property.

And indeed, every Euclidean domain is a PID.

Finally, note that every field F is a Euclidean domain. , we have (where is the multiplicative inverse of b) – and any function can be the norm function.

So, we have a much more interesting chain as follows: {rings} ⊇ {commutative rings (with unity)} ⊇ {integral domains} ⊇ {UFDs} ⊇ {PIDs} ⊇ {Euclidean domains} ⊇ {fields} :-D

2.

3. Thanks for this thread Jane. It is a subject I seemed to have skipped, but have recently been trying to learn in order to please both you and serpicojr.

However, the following doesn't look quite right to me
Originally Posted by JaneBennet
If R is a commutative ring, an ideal in R is a subset I of R such that I is an additive subgroup of R and .
Have you typoed, or have I misunderstood somewhere along the way?

4. Originally Posted by Guitarist
However, the following doesn't look quite right to me
Originally Posted by JaneBennet
If R is a commutative ring, an ideal in R is a subset I of R such that I is an additive subgroup of R and .
Have you typoed, or have I misunderstood somewhere along the way?
Ooh, thanks for pointing that out. It was a typo; I meant

Sorry for the confusion.

I just started sticking my toes into Galois theory, and just want to make sure I know the basic stuff about rings and fields. Iâ€™m pleased you find my post useful.

5. Originally Posted by JaneBennet
Iâ€™ve just started sticking my toes into Galois theory,
Yay! Me too! I'm using, among other more general sources, Ian Stewart's nice little book (I am a huge fan of Stewart's, mainly from his radio broadcasts).

Let's see how we get on, Your throw........

6. I am using A Course in Galois Theory by D.J.H. Garling. The material looks pretty comprehensive. At the moment I am into Chapter 4, on field extensions. 8)

7. Ok, Jane, I just checked field extensions in the sources in front of me.

They look at first glance pretty straightforward, if not slightly trivial. Obviously I am missing something.

Your source is no doubt more comprehensive than mine, So, unless I am too late, hang fire on further tutorials, while I get out the trusty pencil, and try and figure what's going on.

8. Well, the motivation behind field extensions is to solve polynomials equations with coefficients in a given field.

For example, consider , the ring of polynomials with rational coefficients. Not every polynomial equation , where , has a solution in . For instance, has no rational solution. One way to resolve the problem is to find a larger field E which contains as a subfield such that has a solution in E. An obvious candidate is . Unfortunately it turns out that is too large for some purposes. Well, forunately, a smaller field is available: . The task then is to find these so-called algebraic extensions E.

The process starts with as simple a definition as you can get with fields: given two fields E and F, E is said to be an extension of F iff F is a subfield of E. There is nevertheless a good reason for this childishly simple defintion: when we are given a field F, we are often interested not in finding subfields of F but rather in finding bigger fields (i.e. extensions) of which F is a subfield. Hence the motivation in defining extensions of fields. The main trick is to regard an extension E as a vector space over F.

9. OK, I have been thinking about field extensions a bit, and I now acknowledge they are rather subtle and quite interesting.

If is a field, then any field of which is a subfield is said to be an extension field for .My thinking goes like this, see if Jane can agree.

Given the field , it is, at least in principle, relatively straightforward to write down all its subfields. This is because if I will require that includes the additive and multiplicative units, all inverses and satisfies closure under addition and multiplication.

This is quite a constraint!

I will find it helpful to define a field extension for by the field injection . It is usual to identify with its image

(It seems that this may be an inclusion map, or it may not - why not? Dunno)

Anyway, for these injections, the field axioms of units and inverses come free of charge, so these are no constraints on what my extension might be.

Moreover, the axiom of closure is generally mandated by my choice of extension - indeed it seems that closure is often the main motivation for defining the extension. In other words, given a field, I can find (almost) as many extensions as I choose.

Like, the reals are famously not algebraically closed. However, algebraic closure is achieved by the extension , where .

But, we can go the other way too. Let's use the example that Jane gave: consider the rational expression This has the solution , and induces ("generates" I believe is the correct term) the extension .

But this extension is also generated by the expression , whose solution is , which is clearly an element in the field .

Whew, what a wind-bag I am!!

10. Originally Posted by Guitarist
I will find it helpful to define a field extension for by the field injection . It is usual to identify with its image
Yes, this is a much better definition of an extension of a field, actually. is, strictly speaking, generally not a subfield of its extension , but only isomorphic to a subfield of . However, in such situations, abstract algebraists will almost always say that is a subfield of – since abstract algebraists generally don’t make any distinction between isomorphic structures. :P

A note of caution here: be careful not to confuse the terms “extension field”ť and “field extension”ť. An extension field is a field of which your given field is a subfield (or in which can be embedded as a subfield). A field extension, on the other hand, is the dimension of considered as vector space over , and may be denoted . There are many important results with field extensions – e.g. if are fields, then . (Which is something like the field equivalent of Lagrange’s theorem for groups.)

11. Originally Posted by JaneBennet
Finally, note that every field F is a Euclidean domain. , we have (where is the multiplicative inverse of b) â€“ and any function can be the norm function.
Sorry I don't get this part. Why is it that any function can be the norm function of a field?

12. From the definition of a Euclidean domain:

If such that and either r = 0 or , then R is said to satisfy the Euclidean property.
Because any a and b (b ≠ 0) in a field can be written in the above form with r always equal to 0, it does not matter what sort of function you define as your norm (or whether you define any norm at all).

13. I am giving this thread the kiss-of-life mainly on its merits, but also because I have decided to give Galois one more chance (big-hearted me!)

I have a few questions but first a comment.
Originally Posted by JaneBennet
A note of caution here: be careful not to confuse the terms “extension field” and “field extension”.
Well all sources available to me use these terms interchangeably.
. A field extension, on the other hand, is the dimension of considered as vector space over , and may be denoted .
What you describe is what my sources call "the degree of an extension". My comment is not intended to be quarrelsome, as I am fully aware that terminology varies.

Anyhoo, however one frames it, it is a very interesting construction, and one to which I would like to return at some stage.

But first this really dumb question.

Some authors use the term the "roots of a polynomial", some use the term the "zeros" whereas some use them interchangeably. What gives? So, here is my naive take....

We may think of a polynomial over the field as a form , in which case we may speak of the roots of the equality . Here the "x's" are indeterminant.

But we may also have that , where the "x's" are variables for the function , in which case the term "root" has no meaning, and one may use the the term "zero of a function"

In either case, we will have that if, for any , the expression then is alternatively a root or a zero for the polynomial , whether it be thought of as an expression or as a function

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