Fields and Homomorphisms



Definition (Field)

A field   is a commutative unital ring such that every non-zero   has a multiplicative inverse. In other words, for every   there exists some   such that  .

Essentially, a field is a commutative division ring.


  1.   (rational, real and complex numbers) with standard   and   operations have field structure. These are examples with infinite cardinality.
  2.  , the integers modulo   where   is a prime, and   and   are mod   is a family of finite fields.
  3. If   is a field, then  , the set of rational functions (i.e. quotients of polynomials), with coefficients in   also forms a field.
  4. A non-example is   where   is not prime. For example, 2 in   has no multiplicative inverse, hence   is not a field.


Definition (Field Homomorphism)

If   are fields then   is a field homomorphism if


Therefore a field homomorphism is exactly a unital ring homomorphism.

Lemma 4.1.1

Every field homomorphism is injective.

Proof. This is a simple consequence of the ideal structure of fields. Suppose   is a field homomorphism. In particular it is a ring homomorphism so we know that   is a an ideal of  . Since   is a field, it only has trivial ideals so   or  . We can eliminate the second case since   so the map cannot be trivial. Therefore we are in the first case which means exactly that   is injective.       

The above lemma means that every field homomorphism can also be thought of as an embedding of fields.

As happens so often in mathematics, a map between objects induces further maps between related objects. For example, a continuous map between topological spaces induces a map between the set of closed curves on the spaces and a linear map between vector spaces induces a linear map between the dual spaces (albeit in the opposite direction). In this case, a homomorphism between fields induces a homomorphism between the corresponding ring of polynomials. To be precise, suppose   is a field homomorphism. This induces a map   given by 

It is easy to see that   is a (unital) ring homomorphism. Moreover if   is an isomorphism then so is  .

Characteristic of Fields


An important property of fields is their characteristic. We first need to consider the canonical homomorphism   from   into a field  . Of course this is defined by mapping the unit to the unit. Since   is generated by  , this is sufficient to define the entire homomorphism. From the First Isomorphism Theorem, we know that   . In particular, this means that   is a subring and even a subfield of   so is an integral domain. Hence   is a prime ideal of  . There is a unique non-negative integer generating this ideal. We call this integer the characteristic of  . Notice by the above argument that the characteristic must be prime if it is non-zero.

Intuitively, the characteristic of a field   is the smallest positive integer  , if one exists, such that  If no such positive integer exists, then   has characteristic 0. So for example,   all have characteristic   while   and   have characteristic 0.

Sometimes, one calls the image of   under the above canonical homomorphism above the prime subfield of  . Hence the prime subfield of a finite field is (isomorphic to)   (where   is the characteristic of  ) and the prime subfield of a field of characteristic 0 is (isomorphic to)  .

Field Extensions

Definition (Field Extensions)

Let   and   be fields. If   and there is an embedding from   into  , then   is a field extension of  .

Let   be an extension of  . Since we can scale elements of   by elements of   via the multiplication on  ,   forms a vector space over   (one can verify all the axioms for vector spaces hold). The dimension of this vector space is the degree of the extension,  . If the degree is finite, then   is a finite extension of  , and   is of degree   over F.


  • The complex numbers   are a field extension of the real numbers  . The extension is of degree 2.
  • Similarly, one can add the imaginary number   to the field of rational numbers   to form   the field of Gaussian rationals. This is also a degree 2 extension.
  • The real numbers   form a field extension over   but this is not a finite extension since the real numbers do not form a finite dimensional (or even a countably infinite dimensional) vector space over  .

Algebraic Extensions

Definition (Algebraic Extensions)

Let   be an extension of  . Then   is algebraic over   if there exists a non-zero polynomial   such that  .   is an algebraic extension of   if   is an extension of  , such that every element of   is algebraic over  .

For example,   is an algebraic extension over   (if   is any element of  then it is a root of  ) but   is not algebraic over   because for example   is not the root of any rational polynomial (this is a very difficult statement to prove).

Definition (Minimal Polynomial)

If   is algebraic over   then the set of polynomials in   which have   as a root is an ideal of  . This is a principle ideal domain and so the ideal is generated by a unique monic non-zero polynomial,  . We define the   to be the minimal polynomial.

For example, the minimal polynomial of   is   and the minimal polynomial of   is  , both over  . Note the minimal polynomial is heavily reliant on the field it is being viewed over. The minimal polynomial of   over   is simply  .

Splitting Fields


Our primary goal in this study is to find the roots of a given polynomial. The brilliant insight of Galois and Galois theory is to (try to) answer this question by looking at field extensions. The following two lemmas might help motivate this reasoning.

Lemma 4.1.2

Suppose   is a field   is a polynomial. Then there exists a (finite) field extension   of   such that   contains a root of  .

Proof. Suppose first that   is irreducible. Then we can take  . We know that   is indeed a field because   is irreducible. Moreover it contains an isomorphic copy of   as the (equivalence classes of) the constant polynomials. Finally  , the equivalence class of the linear polynomial  , is a root of   since  Finally the degree of   over   is exactly the degree of the polynomial   (which hopefully motivates the terminology). This is due to the division algorithm. Suppose   is any polynomial in  . Then we know by the division algorithm that there exist unique polynomials   and   such that  where  . In particular, this means every equivalence class   contains a unique representative whose degree is less than  . Therefore   is spanned by   where  . If   is not irreducible then it can be written as a product of irreducibles and applying the above process to any of these produces an extension which contains a root of at least one of these irreducible polynomials and hence contains a root of  .     

We know   is irreducible over  , therefore   is a field and one can verify that this field is isomorphic to  . In fact, sometimes one defines the complex numbers as this quotient  .

Lemma 4.1.3

Suppose   is an extension over  . Let   be an irreducible polynomial in   such that   contains  , a root of  . Let   be the smallest subfield of   containing   and  . Then  .

Proof. By the smallest subfield containing   and  , we mean the intersection of all subfields of   that contain them. This collection of fields is non-empty since it contains   for instance and it is easy to see that the intersection of subfields is again a subfield.

If   is of degree 1, then we are done since that would mean that   so   and by the argument towards the end of Lemma 4.1.1, we have  . Then we can assume that  .

In order to show the isomorphism, we define a ring homomorphism  In other words,   acts on polynomials by simply evaluating them at  . By definition, we know that   since  . Since   is irreducible by assumption, it must also then generate the kernel (otherwise it would be a non-trivial multiple of the the generator of the kernel). Then by the First Isomorphism Theorem, we know that   is isomorphic to a subfield of  . Notice that   contains   as the image of the constant polynomials and it contains   as the image of  . By assumption,   was the smallest subfield containing these two things so we must have  .      

The first lemma above tells us that we can always find a field extension containing the root of an irreducible polynomial by modding out by the polynomial. The second lemma tells us that any field extension containing a solution is of this form (up to isomorphism). Thus we will spend considerable time looking at the ring of polynomials over a field and studying its quotient spaces.

One often thinks of   as 'adjoining' the root   to the field  . Roughly speaking, we add   to the field and then we close it under the field operations by also adding in all the possible sums, products, inverses, etc. and the further condition that   satisfies the given polynomial. In fact, this is precisely what the construction in the previous lemma does.

An important consequence of Lemma 4.1.3 is that the roots of an irreducible polynomial are algebraically indistinguishable (this is made precise in Theorem 4.1.4 and in particular by its Corollary 4.1.5). For example, we know that   and   are both solutions of  . There is no algebraic distinction between the two roots; to differentiate them we need topological information like the fact that   and  . Similarly,   and   are both solutions to  . Interchanging these roots is exactly what leads to complex conjugation. The fact that roots of (irreducible) polynomials are all equivalent to one another is one of the key ideas of Galois Theory.

Theorem 4.1.4

Let   be fields and   be an irreducible polynomial. Let   be an isomorphism of fields. Let   be the polynomial  . Let   be a root of   (in some extension of  ) and let   be a root of   (in some extension of  ). Then there exists an isomorphism   that agrees with   on  .

Proof. Since   is an isomorphism and   is irreducible, we must have   is also irreducible (since if we had   then   which would contradict irreducibility of  ). Then   and   generate maximal ideals in their respective rings and the ring isomorphism   descends to an isomorphism (of fields) of the quotients We know by the previous theorem that the domain is isomorphic to   and the codomain is isomorphic to   and this map agrees with   on   by construction.      

Corollary 4.1.5

Let   be fields and   be an irreducible polynomial. Suppose   are roots of   in some (potentially different) extensions of  . Then  .

Proof. Apply the previous theorem to case with   and   as the identity map.

Definition (Splitting Field)

Let   be a field,   and   are roots of  . Then the smallest field extension   of   which contains   is called a splitting field of   over  . In other words, no proper subfield of   contains   and all  .

Existence and Uniqueness of Splitting Fields


We will see that rather than looking at arbitrary field extension, splitting fields will be the things to consider. First we need to know that they always exist.

Theorem 4.1.6

Let   be a field and   a polynomial. Then there exists a field extension   of   that is a splitting field of  .

Proof. This is a largely uninteresting case of proof by induction. We will induct on the degree of  . If   is linear, then clearly its roots (in fact just the one root) is contained in   so   itself is a splitting field. Suppose  . If   splits into the product of linear terms, then again all the roots are contained in  , so we already have a splitting field. So suppose   has an irreducible factor of degree at least 2. Then there exists a field extension   containing a root   of  . Then in  , we can factorise the polynomial into   where   is a polynomial of degree  . Then by induction there exists   a field extension of   that is a splitting field of  . Therefore   is a field extension of   that contains all the roots of  . Taking the intersection of all subfields of   containing   and the roots of   gives us  , a splitting field of  .  

Above we were careful to say a splitting field of  . In fact, this was an unnecessary precaution since the splitting field of a polynomial is unique up to isomorphism. This follows from a generalisation of Theorem 4.1.4, where we claim the statement of the theorem holds even if we adjoin all the roots of the polynomial, instead of just one.

Theorem 4.1.7

Let   be fields and   be a polynomial. Let   be an isomorphism of fields. Let   be the polynomial  . Let   a splitting field of   and   a splitting field of  . Then there exists an isomorphism   that agrees with   on  .

Proof. This is once again a proof by induction on the degree of  . If   is of degree 1 or indeed splits into factors of degree 1 then the splitting field of   is   so we can take  . Thus suppose   has an irreducible factor   of degree at least 2 so  is an irreducible factor of  . Then by the previous theorem we know   extends to an isomorphism   where   is a root of   and   is a root of  . Therefore over   and   respectively we can write   and  . Notice that   is a splitting field of   over  . Indeed if a splitting field was strictly contained within  , then it would contain all the roots of   and   and hence would contain all the roots of  . But this would contradict   being a splitting field of  . Of course the same holds true for   over  . Since   and   have degree strictly less than  , by induction we can assume that the statement of theorem holds for them. In particular,   extends to an isomorphism  . But since   was an extension of  ,   must also be an extension of   concluding the proof.     

Corollary 4.1.8

Let   be a field and   be a polynomial. If   are splitting fields of  , then they are isomorphic.

Proof. Apply Theorem 4.1.7 to the case with   and   as the identity map.      

Classification of Finite Fields

Theorem 4.1.9

If   is a finite field, then   for some prime   and natural number  .

Proof. Since   is a finite field and we know its prime subfield is   for some prime  . The prime subfield is in particular a subfield of   and hence   forms a vector space over  . Since   is finite, it must be a finite dimensional vector space and in particular we must have have   for some   (as vector spaces) so  .      

Theorem (every member of F is a root of  )


let   be a field such that  , then every member is a root of the polynomial  .

proof: Consider   as a the multiplicative group. Then by la grange's theorem  . So multiplying by   gives  , which is true for all  , including  .

Theorem (roots of   are distinct)


Let   be a polynomial in a splitting field   over   then the roots   are distinct.