Abstract Algebra/Fields

Definition
Essentially, a field is a commutative division ring.

Examples
  $$\mathbb{Q}, \mathbb{R}, \mathbb{C}$$ (rational, real and complex numbers) with standard $$+$$ and $$\cdot$$ operations have field structure. These are examples with infinite cardinality.

 $$\mathbb{Z}_p$$, the integers modulo $$p$$ where $$p$$ is a prime, and $$+$$ and $$\cdot$$ are mod $$p,$$ is a family of finite fields. 

 If $$F$$ is a field, then $$F(x)$$, the set of rational functions (i.e. quotients of polynomials), with coefficients in $$F$$ also forms a field. 

 A non-example is $$\mathbb{Z}_n$$ where $$n$$ is not prime. For example, 2 in $$\mathbb{Z}_4$$ has no multiplicative inverse, hence $$\mathbb{Z}_4$$ is not a field.  

Homomorphisms
Therefore a field homomorphism is exactly a unital ring homomorphism.

Proof. This is a simple consequence of the ideal structure of fields. Suppose $$f: E \to F$$ is a field homomorphism. In particular it is a ring homomorphism so we know that $${\rm ker}(f)$$ is a an ideal of $$E$$. Since $$E$$ is a field, it only has trivial ideals so $${\rm ker}(f) = \{0\}$$ or $${\rm ker}(f) = E$$. We can eliminate the second case since $$f(1_E) = 1_F$$ so the map cannot be trivial. Therefore we are in the first case which means exactly that $$f$$ is injective. $$\blacksquare$$

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 $$\varphi: E \to F$$ is a field homomorphism. This induces a map $$\varphi_*: E[x] \to F[x]$$ given by$$\varphi_*(a_0 + a_1x + \dots a_n x^n) = \varphi(a_0) + \varphi(a_1)x + \dots \varphi(a_n)x^n$$

It is easy to see that $$\varphi_*$$ is a (unital) ring homomorphism. Moreover if $$\varphi$$ is an isomorphism then so is $$\varphi_*$$.

Characteristic of Fields
An important property of fields is their characteristic. We first need to consider the canonical homomorphism $$\varphi$$ from $$\mathbb{Z}$$ into a field $$F$$. Of course this is defined by mapping the unit to the unit. Since $$\mathbb{Z}$$ is generated by $$1$$, this is sufficient to define the entire homomorphism. From the First Isomorphism Theorem, we know that $$\mathbb{Z}/\textrm{ker}(\varphi) \cong \varphi(\mathbb{Z})\subseteq F $$. In particular, this means that $$\varphi(\mathbb{Z})$$ is a subring and even a subfield of $$F$$ so is an integral domain. Hence $$\textrm{ker}(\varphi)$$ is a prime ideal of $$\mathbb{Z}$$. There is a unique non-negative integer generating this ideal. We call this integer the characteristic of $$F$$. Notice by the above argument that the characteristic must be prime if it is non-zero.

Intuitively, the characteristic of a field $$F$$ is the smallest positive integer $$p$$, if one exists, such that $$\underbrace{1 + \dots + 1}_{p \text{ times}} = 0$$If no such positive integer exists, then $$F$$ has characteristic 0. So for example, $$\mathbb{Z}_p$$ all have characteristic $$p$$ while $$\mathbb{Q}, \mathbb{R}$$ and $$\mathbb{C}$$ have characteristic 0.

Sometimes, one calls the image of $$\Z$$ under the above canonical homomorphism above the prime subfield of $$F$$. Hence the prime subfield of a finite field is (isomorphic to) $$\mathbb{Z}_p$$ (where $$p$$ is the characteristic of $$F$$) and the prime subfield of a field of characteristic 0 is (isomorphic to) $$\Q$$.

Examples

 * The complex numbers $$\mathbb{C}$$ are a field extension of the real numbers $$\mathbb{R}$$. The extension is of degree 2.
 * Similarly, one can add the imaginary number $$i$$ to the field of rational numbers $$ \mathbb{Q} $$ to form $$\mathbb{Q}(i)$$ the field of Gaussian rationals. This is also a degree 2 extension.
 * The real numbers $$\mathbb{R}$$ form a field extension over $$\mathbb{Q}$$ 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 $$\mathbb{Q}$$.

Algebraic Extensions
For example, $$\mathbb{Q}(\sqrt{2})$$ is an algebraic extension over $$\mathbb{Q}$$ (if $$a + b\sqrt{2}$$ is any element of $$\mathbb{Q}(\sqrt{2})$$then it is a root of $$f(x) = (x - a)^2 - 2b^2$$) but $$\mathbb{R}$$ is not algebraic over $$\mathbb{Q}$$ because for example $$\pi$$ is not the root of any rational polynomial (this is a very difficult statement to prove).

For example, the minimal polynomial of $$i$$ is $$ x^2 + 1$$ and the minimal polynomial of $$\sqrt[3]{2}$$ is $$x^3 - 2$$, both over $$\mathbb{Q}$$. Note the minimal polynomial is heavily reliant on the field it is being viewed over. The minimal polynomial of $$\sqrt[3]{2}$$ over $$\mathbb{R}$$ is simply $$x - \sqrt[3]{2}$$.

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.Proof. Suppose first that $$f(x)$$ is irreducible. Then we can take $$E = F[x]/(f(x))$$. We know that $$E$$ is indeed a field because $$f(x)$$ is irreducible. Moreover it contains an isomorphic copy of $$F$$ as the (equivalence classes of) the constant polynomials. Finally $$[x]$$, the equivalence class of the linear polynomial $$x$$, is a root of $$f(x)$$ since $$f([x]) = [f(x)] = 0 \in F[x]/(f(x))$$Finally the degree of $$E$$ over $$F$$ is exactly the degree of the polynomial $$f(x)$$ (which hopefully motivates the terminology). This is due to the division algorithm. Suppose $$g(x)$$ is any polynomial in $$F[x]$$. Then we know by the division algorithm that there exist unique polynomials $$q(x)$$ and $$r(x)$$ such that $$g(x) = q(x)f(x) + r(x)$$where $$\deg(r(x)) < \deg(f(x))$$. In particular, this means every equivalence class $$[g(x)]$$ contains a unique representative whose degree is less than $$\deg(f(x))$$. Therefore $$E$$ is spanned by $$\{1, [x], [x^2], \dots, [x^{n - 1}]\}$$ where $$n = \deg(f(x))$$. If $$f(x)$$ 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 $$f(x)$$. $$\blacksquare$$

We know $$x^2 + 1$$ is irreducible over $$\mathbb{R}$$, therefore $$\mathbb{R}[x]/(x^2 + 1)$$ is a field and one can verify that this field is isomorphic to $$\mathbb{C}$$. In fact, sometimes one defines the complex numbers as this quotient $$\mathbb{R}[x]/(x^2 + 1)$$.

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

If $$f(x)$$ is of degree 1, then we are done since that would mean that $$\alpha \in F$$ so $$F(\alpha) = F$$ and by the argument towards the end of Lemma 4.1.1, we have $$F[x]/(f(x)) \cong F$$. Then we can assume that $$\deg(f(x)) \geq 2$$.

In order to show the isomorphism, we define a ring homomorphism $$\begin{align} \varphi: F[x] &\to F(\alpha)\\ g(x) &\mapsto g(\alpha) \end{align}$$In other words, $$\varphi$$ acts on polynomials by simply evaluating them at $$\alpha$$. By definition, we know that $$f(x) \in \textrm{ker}(\varphi)$$ since $$f(\alpha) = 0$$. Since $$f(x)$$ 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 $$F[x]/(f(x))$$ is isomorphic to a subfield of $$F(\alpha)$$. Notice that $$F[x]/(f(x))$$ contains $$F$$ as the image of the constant polynomials and it contains $$\alpha$$ as the image of $$x$$. By assumption, $$F(\alpha)$$ was the smallest subfield containing these two things so we must have $$F[x]/(f(x)) \cong F(\alpha)$$. $$\blacksquare$$

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 $$F(\alpha)$$ as 'adjoining' the root $$\alpha$$ to the field $$F$$. Roughly speaking, we add $$\alpha$$ 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 $$\alpha$$ 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 $$\sqrt{2}$$ and $$-\sqrt{2}$$ are both solutions of $$x^2 - 2$$. There is no algebraic distinction between the two roots; to differentiate them we need topological information like the fact that $$-\sqrt{2} < 0$$ and $$\sqrt{2} > 0$$. Similarly, $$i$$ and $$-i$$ are both solutions to $$x^2 + 1$$. 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. Proof. Since $$\varphi$$ is an isomorphism and $$f(x)$$ is irreducible, we must have $$\widetilde{f}(x)$$ is also irreducible (since if we had $$\widetilde{f}(x) = g_1(x) g_2(x)$$ then $$f(x) = (\varphi^{-1})_*(g_1(x)) \cdot (\varphi^{-1})_*(g_2(x))$$ which would contradict irreducibility of $$f(x)$$). Then $$f(x)$$ and $$\widetilde{f}(x)$$ generate maximal ideals in their respective rings and the ring isomorphism $$\varphi_*$$ descends to an isomorphism (of fields) of the quotients$$F[x]/(f(x)) \to \widetilde{F}[x]/(\widetilde{f}(x))$$We know by the previous theorem that the domain is isomorphic to $$F(\alpha)$$ and the codomain is isomorphic to $$\widetilde{F}(\beta)$$ and this map agrees with $$\varphi$$ on $$F$$ by construction. $$\blacksquare$$Proof. Apply the previous theorem to case with $$F = \widetilde{F}$$ and $$\varphi$$ as the identity map.

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.Proof. This is a largely uninteresting case of proof by induction. We will induct on the degree of $$f(x)$$. If $$f(x)$$ is linear, then clearly its roots (in fact just the one root) is contained in $$F$$ so $$F$$ itself is a splitting field. Suppose $$\deg(f(x)) > 1$$. If $$f(x)$$ splits into the product of linear terms, then again all the roots are contained in $$F$$, so we already have a splitting field. So suppose $$f(x)$$ has an irreducible factor of degree at least 2. Then there exists a field extension $$E_1$$ containing a root $$\alpha$$ of $$f(x)$$. Then in $$E_1$$, we can factorise the polynomial into $$f(x) = (x - \alpha)f_1(x)$$ where $$f_1(x)$$ is a polynomial of degree $$\deg(f(x)) - 1$$. Then by induction there exists $$E_2$$ a field extension of $$E_1$$ that is a splitting field of $$f_1(x)$$. Therefore $$E_2$$ is a field extension of $$F$$ that contains all the roots of $$f(x)$$. Taking the intersection of all subfields of $$E_2$$ containing $$F$$ and the roots of $$f(x)$$ gives us $$E$$, a splitting field of $$f(x)$$. $$\blacksquare$$

Above we were careful to say a splitting field of $$f(x)$$. 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.Proof. This is once again a proof by induction on the degree of $$f(x)$$. If $$f(x)$$ is of degree 1 or indeed splits into factors of degree 1 then the splitting field of $$f(x)$$ is $$F$$ so we can take $$\sigma = \varphi$$. Thus suppose $$f(x)$$ has an irreducible factor $$p(x)$$ of degree at least 2 so $$\widetilde{p}(x) = \varphi_*(p(x))$$is an irreducible factor of $$\widetilde{f}(x)$$. Then by the previous theorem we know $$\varphi$$ extends to an isomorphism $$\psi: F(\alpha) \to \widetilde{F}(\beta)$$ where $$\alpha$$ is a root of $$p(x)$$ and $$\beta$$ is a root of $$\widetilde{p}(x)$$. Therefore over $$F(\alpha)$$ and $$\widetilde{F}(\beta)$$ respectively we can write $$f(x) = (x - \alpha) f_1(x)$$ and $$\widetilde{f}(x) = (x - \beta)\widetilde{f}_1(x)$$.

Notice that $$E$$ is a splitting field of $$f_1(x)$$ over $$F(\alpha)$$. Indeed if a splitting field was strictly contained within $$E$$, then it would contain all the roots of $$f_1(x)$$ and $$\alpha$$ and hence would contain all the roots of $$f(x)$$. But this would contradict $$E$$ being a splitting field of $$f(x)$$. Of course the same holds true for $$\widetilde{f}_1(x)$$ over $$\widetilde{F}(\beta)$$. Since $$f_1(x)$$ and $$\widetilde{f}_1(x)$$ have degree strictly less than $$\deg(f(x))$$, by induction we can assume that the statement of theorem holds for them. In particular, $$\psi$$ extends to an isomorphism $$\sigma: E \to \widetilde{E}$$. But since $$\psi$$ was an extension of $$\varphi$$, $$\sigma$$ must also be an extension of $$\varphi$$ concluding the proof. $$\blacksquare$$Proof. Apply Theorem 4.1.7 to the case with $$F = \widetilde{F}$$ and $$\varphi$$ as the identity map. $$\blacksquare$$

Classification of Finite Fields
Proof. Since $$F$$ is a finite field and we know its prime subfield is $$\Z_p$$ for some prime $$p$$. The prime subfield is in particular a subfield of $$F$$ and hence $$F$$ forms a vector space over $$\Z_p$$. Since $$F$$ is finite, it must be a finite dimensional vector space and in particular we must have have $$F \cong (\Z_p)^n$$ for some $$n$$ (as vector spaces) so $$\mid F\mid = p^n$$. $$\blacksquare$$

Theorem (every member of F is a root of $$x^{p^n}-x$$)
let $$F$$ be a field such that $$\left\vert F \right\vert = p^n $$, then every member is a root of the polynomial $$x^{p^n}-x$$.

proof: Consider $$F^*=F/{0}$$ as a the multiplicative group. Then by la grange's theorem $$\forall x \in F^*, x^{p^n-1}=1$$. So multiplying by $$x$$ gives $$x^{p^n}=x$$, which is true for all $$x \in F$$, including $$0$$.

Theorem (roots of $$x^p-x$$ are distinct)
Let $$x^p-x$$ be a polynomial in a splitting field $$E$$ over $$\mathbb{Z} _p$$ then the roots $$a_1,...a_n$$ are distinct.