"Abstract algebra" 08-Domain expansion

1. The vegetarian domain and the single-spread domain 1.1-domain

Domain is a relatively "complete" structure, it has more restrictive conditions, the structure of nature is not very diverse. Now we have a preliminary study of the structure of the domain, the method of course is to expand from small domain to large domain, if \ (f\) is \ (e\) of the subdomain, \ (e\) also called \ (f\) The expansion of the domain or expansion . Expansion of course to start from the simplest domain, we are more familiar with the simple domain what? The simplest infinite field is the rational number field, which is the smallest number field, and any number field contains the Rational number field, and the simplest finite field is the remainder class domain \ (z_p\) of integers under primes \ (p\). Both domains no longer have a true subdomain, we refer to a domain without a real subdomain as a plain , generally written as \ (\triangle\).

So apart from these two well-known domains, are there other areas? Each domain contains the unit \ (e\), the domain generated by \ (e\) is all the element domain, and it is the quotient domain of a build ring, so we can discuss from \ (e\) generation ring \ (Z ' =\{ne\}\). When \ (\text{char}\triangle=\infty\), \ (Z ' \) is isomorphic to the integer ring \ (z\) so that their quotient domain is isomorphic, i.e. \ (\triangle\cong\bbb{q}\). When \ (\text{char}\triangle=p\), as discussed earlier, such rings \ (Z ' \) are isomorphic to the same Yu \ (z_p\), and then have \ (\triangle\cong z_p\). In this sense, only \ (\bbb{q}\) and \ (z_p\) are in the homogeneous domain, and any domain contains and contains only a single domain.

1.2 Single extended domain

With the simplest domain, the next step is to expand the domain, and to study the nature of the newly added elements, as well as the structural characteristics of the extended domain. In f\ \ (e\), take subset \ (s\), \ (f\) to add \ (s\) after the generated expansion of the domain (F (S) \), it is important to note that this definition is always based on the existence of the expansion domain \ (e\). Let's talk about the nature of this expansion, and examine \ (f (s_1) (s_2) \), which is defined as the domain containing \ (f,s_1,s_2\), and \ (f (S_1\cup s_2) \) is the smallest domain containing \ (F,s_1\cup s_2\), so there \ (f (S _1\cup s_2) \subseteq F (s_1) (s_2) \). The same can be pushed to \ (f (s_1) (s_2) \subseteq f (s_1\cup s_2) \), so that the formula (1) is obtained.

\[f (S_1) (s_2) =f (s_2) (s_1) =f (S_1\cup s_2) \tag{1}\]

The above conclusion indicates that the expanding domain \ (F (S) \) is equivalent to the local expansion of finite step, and the order of expansion does not affect the result. The study of local expansion will contribute to the expansion of the whole domain, and in particular we can focus first on \ (| s|=1\) (F (\alpha) \), which are called single-spread domains . By the definition of the domain and the characteristics of the fractional, it is easy to know that the elements in \ (f (\alpha) \) have the format \ (\dfrac{f (\alpha)}{g (\alpha)}\), where \ ({F (\alpha)},{g (\alpha)}\) is a polynomial in \ (f\). All fractions form a single field, but different fractions may point to the same element, and we will proceed from here to study the structure of the single-spread domain.

The polynomial is the infrastructure in the extended domain, and the discussion of it can help us to analyze the structure of the domain. Substituting \ (\alpha\in e\) in \ (f\) for all the polynomial \ (f[x]\), the resulting value may be 22 different, or duplicate may occur. When repetition occurs, subtracting the polynomial will get \ (f (\alpha) =0\), there is such a polynomial of \ (\alpha\) called \ (f\) algebraic element , otherwise known as the transcendental element . There is an essential difference between algebraic element and transcendental element, and it is necessary to discuss the structure of single expanding domain from this angle. For the expansion of the rational number field in the real field, the algebraic number is the algebraic element, and the transcendental number is the transcendental element, which is actually an extended discussion of them.

For many polynomial that satisfies \ (f (\alpha) =0\), you can always find the lowest number of first \ (1\) polynomial. It is easy to prove to algebraic element \ (\alpha\) that this polynomial exists and is unique, it is called \ (\alpha\) on \ (f\) the smallest polynomial \ (p (x) \). The number of minimum polynomial is also referred to as the number of algebraic elements, obviously \ (f\) The number of the element is \ (1\). The minimum polynomial is somewhat simple in nature, first of all it is irreducible on \ (f\), otherwise it must have a factor satisfying \ (g (\alpha) =0\), which contradicts the definition of the minimum polynomial. Secondly, for any polynomial that satisfies \ (f (\alpha) =0\), there must be a \ (P (x) \mid f (x) \), otherwise the polynomial satisfying \ (r (x) =0\) can be constructed with the remainder division.

The structure of the single-spread domain is more obvious around the element type or the least polynomial. Although intuition has already told you the final answer, you still have to use rigorous reasoning to validate the conjecture. The inference method, of course, starts from the definition of appropriate homomorphic mappings, validates the isomorphism of the generated loops, and then deduces the isomorphism of the quotient domain. When \ (\alpha\) is the transcendental element, the generated ring is obviously isomorphic to \ (f[x]\), thus \ (f (\alpha) \) isomorphic to its quotient ring \ (f (x) \). When \ (\alpha\) is an algebraic element, it can be proved that the generation ring \ (f[x]\) is isomorphic to \ (F[x]/\langle p (x) \rangle\), because \ (p (x) \) is irreducible, the expression is a domain, so there \ (F[\alpha]=f (\alpha) \)。 Thus, the single expansion domain of algebraic elements is a polynomial ring (equation (2)) which is modeled by \ (p (x) \), which shows the concise structure of the single algebraic extension and the importance of the study of algebraic expansion.

\[f (\alpha) =f[\alpha]\cong F (x)/\langle p (x) \rangle\tag{2}\]

The above results also indicate that if \ (\alpha\) is the number of \ (n\), then \ (f (\alpha) \) Any element is a number of times less than \ (n\) of the value of the polynomial \ (f (\alpha) =a_0+a_1\alpha+\cdots+a_{n-1} {\alpha}^{n-1}\), in other words, each element is a linear combination of \ (1,\alpha,\cdots,{\alpha}^{n-1}\) on \ (f\), and it is easy to prove that the notation is unique. In the language of linear algebra, the single algebraic extension \ (F (\alpha) \) is a \ (n\) dimension space on \ (f\), and the base of the space is \ (1,\alpha,\cdots,{\alpha}^{n-1}\). It is also useful to analyze the spread of single algebras from this perspective.

2. Algebraic expansion 2.1 Algebraic expansion domain

After figuring out the structure of the single algebraic expansion, we would like to further investigate the expansion of the domain generated by more algebraic elements, or the expansion of all the elements that are algebraic entities. First of all, the natural question is, are these two widening domains the same? For the convenience of the discussion, we define the latter as the algebraic expansion domain, which contains the extended domain of the transcendental element is called the transcendental expansion domain . Since algebraic expansion is always generated by algebraic elements, the question that has just been changed naturally is: is the extended domain generated by the algebraic meta set \ (s\) (F (S) \) necessarily an algebraic extension? The intuition tells us that this conclusion is set up, but it is not so obvious to ponder it carefully. Now we have two steps to prove this speculation, first consider \ (s\) as a finite set of scenes, and then extended to infinity set.

The linear spatial structure of the single algebraic expansion suggests that we study the dimensions of the more general domain, if the expansion \ (e=f (S) \) is a linear space on \ (f\), the dimension of this space is called \ (e\) on \ (f\ ), and is recorded as \ ([e:f]\). \ ([e:f]\) is limited, \ (e\) is called \ (f\) The finite sub-domain , otherwise called infinite spread domain . With the simple deduction of linear algebra, we can get the cumulative number of times (equation (3)). In the case of finite e\, the base \ (a_1,\cdots,a_m\) on \ (k\), \ (k\) on \ (f\) is based on \ (b_1,\cdots,b_n\), easy to prove \ (a_ib_j\) is \ (e\) in \ (f\) The base (expressed in linear and proved irrelevant).

\[[e:f]=[e:k][k:f]\tag{3}\]

For any \ (n\) sub-domain, examine the power of any element \ (1,\alpha,\cdots,\alpha^n\), this \ (n+1\) element must be linearly related, so \ (\alpha\) must be an algebraic element. This means that the finite sub-domain is always an algebraic extension, and it can be used to judge the algebraic expansion domain. On the other hand, when the current set of elements (S\) is a finite set, the \ (f (s) \) can be obtained by a finite number of single algebraic expansions, known by the formula (3) \ (f (s) \) is the finite sub-domain of \ (f\), and thus it is also an algebraic extension. This conclusion directly indicates that the arithmetic of algebraic elements is still an algebraic element. When \ (s\) is an infinite set, the elements in \ (F (S) \) can be represented as a finite arithmetic of elements in \ (F\cup s\), which is also an algebraic element. Combined with the above two points, we get the conclusion that \ (F (S) \) is always algebraic spread.

With this basic conclusion, you can easily prove that the algebraic expansion of the algebraic expansion domain of \ (f\) is still the algebraic extension of \ (f\). If the extended domain \ (e\) is not an algebraic extension, we can take out all the algebraic elements in it, it is easy to prove that they are composed of a set \ (k\) is a domain, and thus \ (K\) is the algebraic extension of \ (f\). \ (k\) is the largest algebraic extension in \ (e\), and the element \ (\alpha\) in \ (E-k\) is the transcendental element, and thus \ (e\) is the pure transcendental domain of \ (k\). In this way, the analysis of any extended domain can be divided into the algebraic expansion and the pure beyond the expansion of the domain of the discussion.

Given the special status of the polynomial, there is a class of algebraic expansions associated with it, which needs to be discussed in particular. The most important nature of the polynomial is its root, the root \ (\alpha\) can be decomposed from \ (f (x) \) The item ((X-\alpha) \), if \ (f (x) \) can be completely decomposed into an item, all its roots represent the polynomial. The domain called algebraic closure of any polynomial can be completely decomposed, and it is obvious that any algebraic extension is itself, it can no longer expand.

The condition of algebraic closure is still too strong for most of the time, perhaps it is more useful to discuss a domain that cannot be expanded for a polynomial. Think of a polynomial \ (f (x) \) on \ (f\), if \ (f (x) \) cannot be fully decomposed, take the irreducible (p (x) \) in it, expand it to the minimum polynomial, and (p (x) \) must be approximate in the expanded domain. So after a finite step, \ (f (x) \) can be fully decomposed in the extended domain \ (e\), and the isomorphic meaning \ (e\) is unique, it is called \ (f (x) \) on \ (f\) on the split domain . The split domain is actually the generated domain \ (f (x_1,\cdots,x_n) \) for all the roots of \ (f (x) \), so it is also called the root domain , and the definition of the split domain makes it easier to discuss the polynomial, and further content will be discussed in the next article.

2.2-foot gauge drawing problem

After introducing the basic concept of expanding domain, we take a look at its application in drawing to exercise the ability to solve practical problems with abstract concepts. Ruler mapping is a traditional method of drawing, which uses simple tools to obtain complex and precise graphics. Even so, history has a number of stubborn drawing problems plaguing people, the classic is called "the three major problems of the classical drawing ." They are: Three equal angles, the circle as a square, the cubic cube, here we use the expanded language to demonstrate that they can not be made by the ruler. Although it has been proved that it is not feasible, but there are still some tireless efforts to try, the establishment of scientific spirit is sometimes more important than diligence.

What is the ruler of the drawing, the general book on the issue is not clear, but it is very important to understand the impossibility of drawing difficulties, the following are some personal understanding. First we assume that the purpose of the plot is to get some exact point, not a straight line or curve, or draw a line or a circle at will, and the amount of the drawing puzzle is actually in it. Second, we want to clarify that all the discussion here is confined to three major problems or similar problems, precisely, the known condition of the drawing is only some segment or angle. Because if there are some auxiliary curves in advance, these problems can be made in fact. Again we have to assume that each step of the drawing is from the fixed point (the end point of the line segment, the angle of three points), not from the line or the angle of the non-point mapping. Some people may have the same question as me, how to look at the arbitrary pick-up point of the situation (such as the midpoint of the line)? I do not make a complete deduction here, but I guess the point can also be used to make those points, concrete argument and when it is a question.

Now look at what the ruler can do: The ruler is used to draw a straight line through two points, the compass can only take a certain point as the center, with a given radius of two points to draw a circle. Since the initial conditions are some points on the plane, you can select two of them as \ (0\) and \ (1\) on the real axis so that all points can see a vector (complex number) on the complex plane, remembering that the set of these complex numbers is \ (b\). Next, according to the previous description, use the ruler to make a definite line and circle, to get more definite points, if all can be made in a finite step set of points (plural) to be recorded as \ (s\), point \ (z\) can be made the necessary and sufficient conditions are \ (Z\in s\).

Based on the knowledge of analytic geometry, we can actually calculate the coordinates of the new point (complex number) from the known point, and by simple verification you can find that the new complex number can always be represented as a combination of the arithmetic, conjugate, or square root of the element in \ (b\) (as an exercise). This means that \ (s\) is contained in a closure of \ (b\) about arithmetic, conjugate, or square root, and conversely it is easy to prove that the arithmetic, conjugate, or square root of any known complex number can be plotted as a ruler (as an exercise), which is a definite definition of \ (s\).

In the following attempt to describe \ (s\) in the language of the expansion, it is easy to know that the rational number can be made, second, the conjugate operation on the arithmetic can be maintained, so you can first define the first extension \ (f_1\) ((4)). In order to be closed on the square root, define the extended domain sequence \ (f_k\) (4), it is easy to prove that any element in \ (s\) will appear in a \ (f_k\) Sooner or later, so there is \ (\cup f_k=s\). Further, a finite Tancovic (5) can actually be inserted between \ (F_k\) and \ (f_{k+1}\). The number of times per spread is \ (1\) or \ (2\), so any number in \ (s\) in \ (f_1\) is the power of \ (2\), which is the necessary and sufficient condition for graphing.

\[F_1=\BBB{Q} (B,\bar{b}), \quad f_{k+1}=f_k (\sqrt{f_k}) \tag{4}\]

\[f_k=k_1\triangleleft k_2\triangleleft \cdots\triangleleft k_n=f_{k+1},\quad k_{i+1}=k_i (\sqrt{a_i}) \tag{5}\]

Now back to the three major drawing puzzles, in which the rounding is square and the cube is given two points, respectively (\sqrt[3]{2}\) and \ (\pi\). In these two problems \ (f_1=\bbb{q}\), can be plotted only in \ (\bbb{q}\) \ (2^n\) times irreducible polynomial root. The smallest polynomial of \ (\sqrt[3]{2}\) is \ (x^3-2\), and Linderman proves that \ (\pi\) is a transcendental number, so none of them can be made. For the tri-equal angle, lift \ (\dfrac{\pi}{3}\) For example, it is given a complex number \ (1+\sqrt{3}i\), easy to have \ (F_1=\bbb{q} (\sqrt{3}i) \). In addition, using the three times-angle formula to know that the complex \ (x_0\) is the root of \ (f (x) =8x^3-6x-1\), and \ (\sqrt{3}i\) is not the root of \ (f (x) \), so there is \ ([F_1 (X_0): f_1]=3\), and thus \ (x_0\) cannot be made. However, it is not said that all corners can not be divided into three, such as \ (\dfrac{\pi}{3},\dfrac{3\pi}{10}\) can be made, please self-verification.

3. Finite field

We have understood the general structure of the domain, and now we need to make further analysis of some common and simple domains that have more special properties. Finite field is a kind of useful domain, which has been widely used in discrete mathematics such as coding. From the previous knowledge we can know that the finite domain \ (f\) is characterized by the prime number \ (P\), \ (f\) contains a prime domain \ (\triangle=z_p\), and it is \ (\triangle\) The finite sub-domain, if set \ ([f:\triangle]=n\), then \ (f\) a common \ (p^n\) element. These are the more intuitive features of finite fields, which are sometimes called Gamma Rovavic , which are recorded as \ (GF (p^n) \). Now there are two more natural questions: \ (p^n\) The domain of the order must exist? Is it unique in the homogeneous sense? It will be analyzed below.

The biggest difference between a domain and a ring is that the domain forms a group on multiplication, which is the root cause of many structural characteristics of the domain. Especially in finite fields, non-0 elements form a finite group, so that not 0 elements satisfy \ (A^{q-1}=1, (q=p^n) \), and any element satisfies \ (a^q=a\). Because these \ (p^n\) elements are different, they are the root of the polynomial \ (f (x) =x^q-x\), which is the split domain of \ (f (x) \) on \ (z_p\). We already know the uniqueness of the split domain, so \ (p^n\) is unique in the homogeneous sense.

The above discussion also inspires proof of existence, on polynomial \ (f (x) =x^q-x\), which sets its division domain on \ (z_p\) to be \ (f\). In \ (f\), examine \ (f (x) \) Any two root \ (\alpha,\beta\), easy to verify \ (\alpha-\beta,\dfrac{\alpha}{\beta}\) is also the root of \ (f (x) \), so that all the roots form a domain. On the other hand, =-1\ (f ' (x)), so that \ (f (x) \) does not have a root, so the root of the domain has \ (p^n\) elements, which proves the existence of the \ (p^n\) Order domain.

Further discussion of the multiplication group of the domain, with a non-0 element on the multiplication of the largest order of \ (m\), first obviously has \ (M\mid q-1\). Secondly, we already know in group theory that the order of any element is a factor of \ (m\), so that they all satisfy the root of \ (f (x) =x^m-1=0\). To make \ (f (x) \) have \ (q-1\) a different root, at least \ (M\ge q-1\), so there is \ (m=q-1\). This conclusion shows that the non-0 element is a cyclic group in multiplication, so that \ (\alpha\) is the element of the order (q-1\), it is easy to prove that the domain is \ (\alpha\) generated on the \ (z_p\) of the single spread domain (formula (6)), \ (\alpha\) is called the domain The original root .

\[GF (p^n) =\triangle (\alpha) \tag{6}\]

Now look at the finite domain \ (f\) which subdomains, the first subdomain of the order must be \ (p^m, (m\leqslant n) \), in the multiplication group also has \ (P^m-1\mid p^n-1\), by the knowledge of elementary number theory has \ (M\mid n\). This conclusion can also be proved by the number of expansions, because \ ([GF (p^n): z_p]=n\), and \ ([GF (p^m): z_p]=m\), so there is obviously \ (M\mid n\). Conversely, when \ (M\mid n\), we need to verify that the \ (p^m\) Order subdomain exists. In fact, the previous proof has given the idea, because \ (P^m-1\mid p^n-1\), so \ ((x^s-x) \mid (x^q-x) \), thus \ (x^s-x\) in the \ (f\) can be completely decomposed, \ (s\) a different root domain is to find the subdomain. This proves that \ (f\) the necessary and sufficient conditions of the subdomain is \ (M\mid n\), in fact, ((x^s-x) \) is \ ((x^q-x) \) The factor, obviously \ (p^m\) Order subdomain is also unique.

　　• from the original root, the structure and elements of the finite domain and its subdomains are discussed.

4. Can be away from the expansion of the domain 4.1 can be out of the yuan

As you can see, the finite field is always a single domain, and the simple structure of the single-spread domain is our favorite, which begs the question: what kind of expansion domain is single-spread? This problem is more difficult to answer, but we can give a class of common algebraic expansion, it is always single-domain. There is a class of irreducible polynomial that does not have a root in its split domain, which is useful for discussing single-spread domains. For this reason, the algebraic elements that define the minimum polynomial without a root (in its split domain) are the removable elements, each element is an extended domain of the detached element, which is called the extensible domain , and the polynomial with no root of the irreducible formula is called the polynomial. Before I can prove the conclusion of the single expansion, let me briefly discuss the properties of the isolated and the isolated domains, which of course should be studied from the irreducible polynomial without the root.

Set \ (P (x) \) is an irreducible polynomial on \ (f\), (P (x) \) and \ (P ' (x) \) Expressions such as formulas (7). \ (P (x) \) The necessary and sufficient condition for having a root is \ (d (x) = (p (x), P ' (x)) \) is greater than \ (0\), by \ (p (x) \) not known \ (d (x) =ap (x), (a\in F) \), and then by \ (d (x) \mid p ' (x) \) get \ (P ' (x) =0\), i.e. \ (a_1=2a_2=\cdots=na_n=0\). Domain features only \ (\infty\) and \ (p\), when \ (\text{char}f=\infty\), can only have \ (a_1=a_2=\cdots=a_n=0\), which with \ (p (x) \) irreducible contradiction, Therefore, the irreducible polynomial of this domain does not have a heavy root. When \ (\text{char}f=p\), can get except \ (K\mid p\) have \ (a_k=0\), so \ (p (x) \) has form \ (g (x^p) \), this is the irreducible polynomial has a root requirement.

\[p (x) =a_0+a_1x+\cdots+a_nx^n,\quad p ' (x) =a_1+2a_2x+\cdots+na_nx^{n-1},\quad a_k\in f\tag{7}\]

With this conclusion, we can continue to study the characteristics of the available elements. Since the irreducible polynomial of a domain characterized by \ (\infty\) does not have a root, all its algebraic elements are non-detached, and all algebraic expansions are off-field. We now only need to study the domain \ (f\) characterized by \ (p\), and set \ (\alpha\) is the of \ (f\), and \ (\alpha\) is of course also \ (f\) on any extension of the free meta-domain. Examine the polynomial (8), which is a polynomial on the expanded domain \ (f (\alpha^p) \), then \ (\alpha\) satisfies the minimum polynomial on \ (f (\alpha^p) \) \ (P (x) \mid F (x) \). Consider that \ (\alpha\) is also \ (f (\alpha^p) \) on the removable element, it must have a \ (p (x) =x-\alpha\), this will get \ (\alpha\in f (\alpha^p) \), thus \ (f (\alpha) \subseteq F (\alpha^p) \). \ (f (\alpha^p) \subseteq f (\alpha) \) is obvious, so there is a conclusion \ (f (\alpha) = f (\alpha^p) \).

\[f (x) = (X-\alpha) ^p=x^p-\alpha^p\tag{8}\]

Conversely, if \ (\alpha\) is a non-detached element of \ (f\), its smallest polynomial has the form \ (g (x^p) \). It is easy to prove that \ (g (x) \) is also an irreducible polynomial on \ (f\), and \ (g (\alpha^p) =0\), so \ (g (x) \) is the minimum polynomial of \ (\alpha^p\) on \ (f\). The number of \ (g (x) \) and \ (g (x^p) \) is significantly different, so \ (f (\alpha^p) \) and \ (f (\alpha) \) are not identical. Both positive and negative proofs have been obtained: \ (\alpha\) is \ (f\) on the necessary and sufficient conditions for the existence of the formula (9) is established, the conclusion of the following discussion will be useful.

\[f (\alpha) = F (\alpha^p) \tag{9}\]

There is a basic question is, can the arithmetic of the yuan or the free yuan? or equivalent proposition: if \ (\alpha,\beta\) is \ (f\) on the removable element, \ (F (\alpha,\beta) \) is the extensible domain? Consider the latter proposition, that is, Q \ (\gamma\in F (\alpha,\beta) \) is a \ (f\) of the free meta? First \ (\gamma\) is of course \ (f (\alpha,\beta) =f (\alpha) (\beta) \) on the removable element, if you want to verify our conjecture, you can first prove a more general proposition: if \ (\alpha\) is the domain of (f (\beta) \) , then \ (\alpha\) is also a removable element on \ (f\). can be isolated in the expansion of the domain of course also can be divorced, this proposition is to ask this transitivity under certain conditions can be reversed? For the \text{char}f=\infty\ scenario, this series of conclusions is clearly set, and the following discussion will only target \ (\text{char}f=p\) domains.

Because \ (\alpha\) is a \beta on \ (f () \), known by the previous conclusion (F (\alpha,\beta) =f (\alpha^p,\beta) \), and we try to prove \ (f (\alpha) =f (\alpha^p) \). You can continue to push this conjecture forward, because \ (f (\alpha^p) \subseteq f (\alpha) \) and \ (f (\alpha) \) is the expansion of \ (f (\alpha^p) \), so the conclusion of the proof is equivalent to \ ([F (\alpha): f (\ alpha^p) (]=1\), which is then equivalent to the formula (10). Finally this proposition is actually to discuss \ (\beta\) in \ (f (\alpha) \) and \ (f (\alpha^p) \) on the minimum polynomial \ (h (x), g (x) \) The number of equal, and obviously have a \ (h (x) \mid g (x) \), it is only required (g (X) \ Mid h (x) \). Similar to the previous method, it is easy to prove that \ (h^p (x) \) is a polynomial on \ (F (\alpha^p) \), and obviously \ (H^p (\beta) =0\), plus \ (g (x) \) No root, can only be \ (g (x) \mid h (x) \). This proves our conjecture, as well as all the inferences, that the arithmetic of the yuan can be divorced from the yuan.

\[[f (\alpha,\beta): F (\alpha)]=[f (\alpha^p,\beta): F (\alpha^p)]\tag{10}\]

4.2 Free and full domain

Now is the time to discuss the relationship between the extended domain and the single domain, we have known that finite fields must be single-domain, and now only need to study the infinite domain. Further, we need to limit the spread domain in the limited domain, and it is easy to know by inductive method, just prove that \ (F (\alpha,\beta) \) is Tancovic (\ (\alpha,\beta\) is \ (f\) The separation element), then any finite sub-domain is a single spread domain. to permit \ (F (\alpha,\beta) \) is a single-spread domain, we need to find the single-domain generator \ (\theta\), which consists of \ (\alpha,\beta\) and \ (f\) elements and can be used to represent \ (\alpha,\beta\). There are many possibilities for such constructs, but we can simply construct one, take \ (\theta=\alpha+k\beta, (k\in f) \), by \ (\alpha=\theta-k\beta\) know only proof \ (\beta\in F (\theta) \).

The minimum polynomial on \ (\alpha,\beta\) on \ (f\) is \ (P (x), q (x) \), because \ (f (\theta) \subseteq f (\alpha,\beta) \), then \ (q (x) \in F (\theta) [x]\ ), to permit \ (\beta\in f (\theta) \), simply construct a polynomial (h (x) \) with \ (q (x) \) Only one common root \ (\beta\) on \ (f (\theta) \). This is required by the use of \ (p (x) \), in order to make \ (h (\beta) =0\), natural can make \ (h (x) =p (\THETA-KX) \). To make \ (h (x) \) not contain the other root \ (q (x) \) \ (\beta_i\), also require \ (\theta-k\beta_i\) not equal to \ (p (x) \) any root \ (\alpha_j\), because \ (\beta_i,\alpha_j\) is limited, it is possible to select \ (k\) in \ (f\) to satisfy the condition. This constructs the \ (\theta\) that satisfies the condition (F (\theta) =f (\alpha,\beta) \).

Just now we have proved that the finite-time separation and expansion domain must be single-spread domain, and some criteria for separating and expanding domain are given. In fact, there are some commonly used domains, their expansion is the separation of the domain, which makes the discussion more simple, for this we define such a domain is a full domain or a complete domain . As already known \ (\text{char}f=\infty\) domain is the full domain, now to study the full domain of \ (\text{char}f=p\) the necessary and sufficient conditions. Full domain requirements do not exist in the form of \ (g (x^p) \) Irreducible polynomial, it is easy to see if \ (g (x) \) The coefficients are \ (a_k^p\) Form, \ (g (x^p) \) must be \ (h^p (x) \) Form, thereby \ (g (x^p) \) can be about. That is, if each element of \ (f\) is a power of \ (p\) of an element, \ (f\) must be a full domain.

Conversely, if \ (f\) is a full domain, there is always \ (\beta\) satisfied \ (f (\beta) =0\) in the split domain of any element \ (\alpha\), \ (f (x) =x^p-\alpha\). and \ (f (x) = (X-\beta) ^p\), so \ (\beta\) on \ (f\) the smallest polynomial can only be \ (x-\beta\), which means \ (\beta\in f\), that is, any element is proved to be an element of the \ (p\) power. To synthesize this two-point analysis, the full domain of \ (\text{char}f=p\) is a sufficient and necessary condition that any element is a power of \ (p\) of an element.

• Verification: Finite fields are full domains.

"Abstract algebra" 08-Domain expansion