**1. Isomorphisms **

We return here to the quotient ring construction. Recall that if is a commutative ring with identity and is an ideal of then is also a commutative ring with identity. Here, where for the equivalence relation defined by iff

Since is an equivalence relation, we have that if and if In particular, any two classes are either the same or else they are disjoint.

In case for some field then is principal, so for some i.e., given any polynomial iff and, more generally, (or, equivalently, or, equivalently, ) iff

In this case, contains zero divisors if is nonconstant but not irreducible.

If is 0,

If is constant but nonzero, then

Finally, we want to examine what happens when is irreducible. From now on suppose that this is the case.

Let be any polynomial in If then Otherwise, since is irreducible. So there are polynomials with Therefore It follows that every nonzero element of is invertible, i.e, we have the following:

Lemma 1is a field whenever is irreducible.

Suppose that If then Since and is a constant, we must have i.e., This means that we can identify each with the corresponding element of and we have:

Lemma 2Suppose that is irreducible. Under the identification for we have that is a field extension of

Whenever convenient, we will just write instead of for

Now we connect these field extensions with algebraic numbers. See last lecture for the definition of the minimal polynomial of an algebraic number and its basic properties.

Theorem 3Let where is irreducible and monic. Let Then is algebraic over with minimal polynomial Moreover,

*Proof:* Let for some with (since is irreducible) and (so ).

Note first that Using our identification of elements of with the corresponding classes in we can write

Since we have This shows that is algebraic over

Since any element of is for some and the same argument as above shows that we obviously have

Let be the minimal polynomial of over Since then Since is irreducible, we must have If is also monic, then we must in fact have

Suppose for example that is algebraic (over ) with minimal polynomial By the above, both and are field extensions of of degree where has a root ( in one case, in the other). Although on the surface both extensions look very different, as far as field theory is concerned, they are indistinguishable:

Theorem 4Suppose that is a field, and that and are two field extensions of (not necessarily subfields of a common larger extension field).If and are algebraic over and then and are isomorphic. Moreover, if then the map defined as follows is an isomorphism: Any element of can be written in a unique way as where the numbers are in Set

*Proof:* As shown in Corollary 8 from last lecture, any element of can be written as stated and, similarly, any element of also has the form for some of degree at most Moreover, such representation is unique. It follows that is well-defined, injective, and surjective. It is straightforward to verify that for any and that

It takes a little bit more effort to check that Once this is done, it follows that is an isomorphism and we are done.

Let be polynomials in such that and We can find polynomials such that and Moreover, such polynomials are unique.

It follows that and therefore

Since and then

as needed.

The converse of the previous result is also true. Before stating this precisely, we need a general result about injective homomorphisms.

Lemma 5Let and be two field extensions of (not necessarily subfields of a common larger extension field). Suppose that is an injective homomorphism and that fixes all the elements of i.e., for all If is a subset of linearly independent over then is a subset of linearly independent over

*Proof:* Suppose Let and suppose that

Since the above equals Also, Since is 1-1, it follows that

Since the are linearly independent over then This shows that is linearly independent over

Corollary 6If and are field extensions of and is an isomorphism that fixes all the elements of then

*Proof:* The lemma shows that if is a basis of over then is linearly independent over Since then

The map is an isomorphism from to and it fixes all the elements of The same argument as above implies that

The result follows.

Theorem 7Suppose that is algebraic over that and that is an isomorphism that fixes all the elements of Then and is as given in Theorem 4. Moreover,

*Proof:* By the corollary, say, where is the minimal polynomial of over

We know that is a basis of over The corollary also shows that is a basis of over This gives that and that is as in Theorem 4, i.e.,

Moreover, so is a root of Since is irreducible,

**2. Examples **

The results above give us a very detailed control over the isomorphisms that we can expect to have between fields.

**1. **For example, it follows that is isomorphic to precisely three subfields of namely:

where, as usual,

To see this, note that Theorems 4 and 7 give us that if then is isomorphic to iff for some root of This gives that the only subfields of isomorphic to are the ones listed above. That there are three of them is the same as saying that they are all different.

This can be checked as follows: First, and since while the other fields contain complex nonreal numbers. Second, since otherwise

Now, has degree 2, so would have to divide by the tower law:

whenever This is of course impossible since

**2. **Continuing with this example, we can now easily identify all the subfields of where We know that Since by the tower law it follows that

and that any (strictly) intermediate field between and is a field extension of of degree either 2 or 3.

Let be such a field. Suppose first that If is irreducible over then

and

which, by the tower law, would imply that contradiction.

It follows that is not irreducible over and therefore it has a root in from which it follows that This is because so if it factors, one of its factors is linear, so we must have a root (and by the tower law).

Thus, the only subfields of of degree 3 over are

Suppose now that If is irreducible over then

and

which again leads to a contradiction with the tower law, since

It follows that is not irreducible over and therefore it factors over and

**3. **Suppose now that is a prime and is a finite field of characteristic Then, in particular, so is a finite extension of If then it follows that To see this, suppose that is a basis. Then all the combinations

with are different, and every element of is expressible in this form. To count the number of these expressions, note that there are possible values for each

For example, if to build a field of size 4, it would suffice to find an irreducible polynomial of degree 2, and to form The only candidates are The first and third are not irreducible since 0 is a root. The second is not irreducible since The last one is irreducible, and it is our only choice.

Let now so The elements of are then To write the multiplication table, only needs to be determined. For this, note that

**4. **For a slightly less trivial example, let’s now build a field of 8 elements. Then we need an irreducible polynomial of degree 3. The candidates are:

- none of which are irreducible since is a root,
- which are not irreducible since is a root,
- or Both these polynomials are irreducible.

We have therefore two possible fields of size 8: and

*Case 1: *

Let Then the elements of are and To write the multiplication table, we simply have to “reduce the corresponding polynomials modulo ” For example, Similarly, we find:

*Case 2: .*

Let Just as before, the elements of are and To write the multiplication table, now we reduce modulo We have:

On the surface, and look rather different. However, notice that if we set then In effect, is irreducible and But also, since

Since we can apply Theorem 4 and conclude that and are isomorphic, and that the map given by

for is an isomorphism. In other words, and are actually the same field, but presented in two slightly different ways; what we call in one case, we call in the other.

Notice also that we have that completely factors in Then and is also a root of Similarly, is a root of so in

Let Then also factors in Equivalently, factors in Exactly as above, we find in Equivalently, in Notice that all the `new’ elements of are roots of either or and none of them is a double root (we say that they are **simple** roots) or a root of both and

Finally, note that in so which means that and that every element of is a simple root of the polynomial This is a particular case of a general phenomenon present in all finite fields.

**5. **To illustrate the same phenomenon, note that in the field of 4 elements from example 3, we have that every element is a simple root of the polynomial In general, if is a finite field of size then every element of is a root of so where

I’ll leave it as an exercise to verify that this is the case with the field of 9 elements, and that, in particular, there is only one such field, up to isomorphism. Note that if then is an extension of of degree 2, so one way of building such a field is by considering where is irreducible of degree 3.

There are three such polynomials, namely and One can check that the three extensions one obtains are actually “the same” (i.e., they are isomorphic), and a direct computation verifies that

See also this answer in math.stackexchange.

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