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 1 is 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 2 Suppose 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 3 Let 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:
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
The converse of the previous result is also true. Before stating this precisely, we need a general result about injective homomorphisms.
Lemma 5 Let 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 6 If 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 7 Suppose 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,
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
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
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
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.
Typeset using LaTeX2WP. Here is a printable version of this post.