These are notes to pp. 1-11 and 21-49 of Artin’s classic—”the last corrected printing of the 1944 second revised edition,” republished by Dover in 1998—to assist the study of his account of Galois theory as far as the Fundamental Theorem. In some places I have followed Artin’s revised German edition (Teubner, 1959), which seems not to have been translated into English. I hope that any mistakes that may be noticed will be brought to my attention.


Format

page or paragraph/line
para = paragraph
n = n from bottom
para n = the nth paragraph that begins on the page

1 para 2/-2        is easily seen to be
4 para 2/2         called (linearly) dependent if
4 para 2/-1        called (linearly) independent.
8/-7                  Theorem 2
8/-4                   last m-r
22/-2                 [To agree with later pages, a better notation is:]

a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0

23/1                   [With the “better notation” the condition is:] a_n\ne 0
23 para 2/-3       less than that of f(x)
25 para 1/4        there are non-zero polynomials with this property.
25 para 1/5        non-zero polynomials
25 para 2/-1       [Add at the end of the para:]

The polynomial f(x) is called the minimal polynomial of \alpha over F.

26 para 1/3        [“Finally” should start a new para.]
26 para 2/1        We are about to show that
26 para 2/-3       in consequence E_0
26/-4                 polynomials in F
27/7, 8               the ordinary product
27/12                 exceed n-1 in degree
28 para 1/12      [Delete “of” at the end of the line.]
29/3                   is a field—an extension field of F in which f(x) has a root, namely \xi.
29 para 2/4       [The following clarifies the text from the sentence beginning “This mapping” through para 4:]

Let \phi \colon K \to L be a mapping of fields. For any \alpha in K, \phi (\alpha) is the image of \alpha under \phi. The mapping \phi is a homomorphism if for all \alpha and \beta in K,

(1)   \phi (\alpha + \beta) = \phi (\alpha) + \phi (\beta) and
(2)   \phi (\alpha \beta) = \phi (\alpha) \phi (\beta).

Then \phi (0) = \phi(0+0) = \phi(0)+\phi(0) implies  \phi(0)=0, and then also 0=\phi(0)=\phi(\alpha+(-\alpha))=\phi(\alpha)+\phi(-\alpha) implies

(3)   \phi(-\alpha)=-\phi(\alpha).

Further, \phi(1)=\phi(1\cdot 1)=\phi(1)\phi(1) implies that either \phi(1)=0 or \phi(1)=1.

In the former case, for all \alpha in K, \phi(\alpha) = \phi(1\cdot \alpha) = \phi(1)\phi(\alpha) = 0: the mapping \phi is identically zero.

In the latter case, for all \alpha \ne 0 in K, 1=\phi(1)=\phi(\alpha ^{-1}\alpha) = \phi(\alpha^{-1})\phi(\alpha) implies

(4)   \phi(\alpha) \ne 0 and \phi(\alpha^{-1}) = \phi(\alpha)^{-1}.

The homomorphism \phi is an isomorphism if it is both one-to-one (that is, it sends distinct elements of K to distinct elements of L) and onto (that is, every element of L is the image of an element of K). In that case there is defined the inverse mapping \phi^{-1} \colon L \to K, which takes the image \alpha' = \phi (\alpha) of any element \alpha back to that element: \phi^{-1}(\alpha') = \alpha. This mapping too is an isomorphism, as is easily verified. Fields that admit an isomorphism between them are called isomorphic.

An important special case is that in which L=K, so that the isomorphism is from K to itself, \phi \colon K \to K; it is then called an automorphism of K. Every field admits the identity automorphism, which takes each element to itself.

For the general homomorphism \phi let \phi(K) denote the image of K under \phi; that is, the set of all images \phi(\alpha) of elements \alpha of K. If \phi is onto, \phi(K) = L.

Lemma h.  Let \phi \colon K \to L be a non-zero homomorphism of fields. Then \phi(K) is a subfield of L, and \phi defines an isomorphism of fields K \to \phi(K).

Proof.  If \alpha_1 and \alpha_2 are distinct elements of K, \alpha_1 -\alpha_2 \ne 0, hence by (4) above, and (1) and (3), 0 \ne \phi(\alpha_1 - \alpha_2) = \phi(\alpha_1) - \phi(\alpha_2), so \phi(\alpha_1) and \phi(\alpha_2) are distinct elements of L. Thus \phi is one-to-one, and it is obviously onto \phi(K). That \phi(K) is a subfield of L follows from (1)-(4) above, which show that it is closed under the arithmetic operations in L.

The mapping that takes the element g(\xi) of E_1 to the element g(\alpha) of E_0 is a non-zero homomorphism E_1 \to E, whose image is E_0. By Lemma h, E_0 is a field isomorphic under that mapping to E_1.

29/-2                 a non-constant polynomial
30 para 2/-1      [Add to the end of the sentence:] under which \beta' is the image of \beta.
30 para 3/1       [Replace the proof by the following (after the revised German edition):]

Proof.   Each element \theta of E has the form \theta = g(\beta), where g(x) is a polynomial in F of degree less than that of f(x). Map the element \theta to the element g'(\beta'), where g'(x) is the polynomial in F' that corresponds to g(x). This mapping is obviously an extension of the given mapping \sigma, and maps E onto E'. By Lemma h, we have only to verify that the extended mapping is a homomorphism. Clearly it sends the sum of two elements to the sum of their images. Likewise for the product, because g(\beta) h(\beta) = r(\beta) means that r(x) is the remainder of g(x) h(x) upon division by f(x); i.e., there is a polynomial q(x) such that g(x)h(x) = q(x)f(x) + r(x) (p. 26). Passing to the images of the polynomials gives g'(x)h'(x) = q'(x)f'(x) + r'(x), and putting x = \beta' we have g'(\beta')h'(\beta') = r'(\beta').

30 para 6/1        since (as we are about to show) it is
31/3                  a field D in
31/4                  the field within D generated
31/5                  splitting field E.
31 para 3/1        [Label the first sentence:] (*)
31 para 3/-1       [Add at the end of the para:] This proves (*).
32 para 1/9        corresponding
32 para 2/1        [Insert at the beginning of the para:]

Applying (*), p. 31, let \alpha in E be a root of f_1(x), \alpha' in E' a root of f_1'(x).

34/4                  If (supposing r \le s) we
34 para 1/4        \sigma(\alpha) \cdot \sigma(\beta)
34 para 1/-1       [Add at the end of the para:]

It is easily seen that \sigma maps the identity element of G to 1, and maps the inverse of any element to the multiplicative inverse of its image: \sigma(\alpha^{-1}) = \sigma(\alpha)^{-1}.

36 para 3/3         \sigma_i(a \cdot b) =
36 para 3/4-5      [Replace what follows “we have” to the end of the sentence by:]

\sigma_i (\alpha^{-1}) = \sigma_i (\alpha)^{-1} = \sigma_j (\alpha)^{-1} = \sigma_j (\alpha^{-1}).

37/10                [Insert a comma as shown:] \sigma_1(a_i) = \sigma_j(a_i), and
37/-7                 [After the equation insert a comma, followed by:] valid for every \alpha.
37/-4                 of a field
38/5-6               [Replace what follows “we shall” to the end of the sentence by the following (although it is rendered unnecessary by the definition of inverse already given at 29 para 2/4):]

denote by \sigma^{-1} the inverse of \sigma (the mapping of y into x, i.e. \sigma^{-1}(y) = x).

38/-1                 and call the
38 para 3           [After this para insert the following:]

We now introduce a useful notion, which will soon have application.

If G = \{g_1, g_2, \dots, g_n\} is any finite group, it is easy to see that each of the following two sets of n elements is the same as G: (i) \{gg_1,g g_2, \dots, gg_n\}, where g is any given element of G; (ii) \{g_1^{-1}, g_2^{-1}, \dots, g_n^{-1}\}. Now let G = \{\sigma_1, \sigma_2, \dots, \sigma_n\} be a group of automorphisms of a field E, and let F be the fixed field of G. For any element \alpha in E, the sum

T(\alpha) = \sigma_1(\alpha) + \sigma_2(\alpha) + \cdots + \sigma_n(\alpha)

of the images of \alpha under G is called the trace of \alpha. In view of (i), for each i \{\sigma_i \sigma_1, \sigma_i  \sigma_2, \dots, \sigma_i  \sigma_n\} = G, hence \sigma_i (T(\alpha)) = T(\alpha). Thus the trace T(\alpha) is fixed under G, so it lies in F. Further, because the \sigma_i are independent (Theorem 12, Corollary), there exists \alpha such that T(\alpha) \ne 0: the trace mapping T \colon E \to F is not identically zero. Finally, we note that since in view of (ii)  \{\sigma_1^{-1}, \sigma_2^{-1}, \dots, \sigma_n^{-1}\} = G, the trace can also be written

T(\alpha) =  \sigma_1^{-1}(\alpha) + \sigma_2^{-1}(\alpha) + \cdots + \sigma_n^{-1}(\alpha).

39/1                   fixed field F.
39/6                   seen. Indeed, I(x) = \frac{1}{2} (T(x^2) + 3), where T denotes the trace. Hence
39 para 2/1         from the Corollary to Theorem 13

39 para 2/-1       [The second term of the equation should have I, not 1:] -I \cdot x^2 (x-1)^2
39/-4                 fixed field
39/-3                 The Corollary to Theorem 13
40/6                   succeed
40 para 1/6        S_{i-1} over S_i
41/1                  polynomials in x_2^1, x_3, \dots x_n,
41/2                  polynomials in x_3^2, x_3^1, x_4, x_5,  \dots x_n,
42 para 2/-2      [For a neater continuation of the proof (after the revised German edition), replace from “We note” to the end of the proof (at 43/13) by the following:]

By renumbering the x_i if necessary, we can assume that x_1 \ne 0. We can further arrange it that the trace T(x_1) \ne 0. For if T(x_1) = 0, let \alpha be such that T(\alpha) \ne 0 (p. 38), and multiply the equations by \alpha x_1^{-1}, thus making \alpha the new x_1. Now for each i we apply \sigma_i^{-1} to the ith equation, obtaining

\sigma_i^{-1} (x_1) \alpha_1 + \sigma_i^{-1} (x_2) \alpha_2 + \cdots + \sigma_i^{-1} (x_{n+1}) \alpha_{n+1} = 0.

Adding these equations gives

T (x_1) \alpha_1 + T (x_2) \alpha_2 + \cdots + T (x_{n+1}) \alpha_{n+1} = 0,

because for each j, \sigma_1^{-1} (x_j) + \sigma_2^{-1} (x_j)  + \cdots + \sigma_n^{-1} (x_j) = T(x_j) (p. 38). Since each T(x_j) is in F and T(x_1) \ne 0, this contradicts the assumption that the \alpha_j are linearly independent over F.

44 para 4/1        in E that are not in F.
44 para 5/4        [At the end of the line:] deg p_1(x) = s.
44/-2                 [Delete “a” at the end of the line.]
45/-3                 [The text from “We first prove” to the end of the proof of the lemma (at 46 para 1/-1) is here restated. (The lemma deserves to be a theorem, as it is in the revised German edition mostly followed here.)]

A lemma is needed, but we first make a general observation. Let E be any extension of a field F, \sigma an automorphism of E that leaves F fixed, and f(x) a polynomial in F.

(*)  If \alpha in E is a root of f(x), \sigma (\alpha) is also a root of f(x).

For f(\sigma (\alpha)) = \sigma (f(\alpha)) = \sigma (0) =0.

Lemma.  Let E be a normal extension of F with automorphism group G. Then E is a separable extension of F. More precisely: if \alpha is any element of E, and \alpha_1, \alpha_2, \dots, \alpha_r are the distinct images of \alpha under the automorphisms of G, then

f(x) = (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_r)

is an irreducible polynomial in F with root \alpha; thus f(x) is the minimal polynomial of \alpha over F.

Proof.  Let the automorphisms of G be denoted \sigma_1, \sigma_2, \dots, so that \alpha_1, \alpha_2, \dots, \alpha_r are the distinct elements among the \sigma_i (\alpha); \alpha is one of them. As we remarked in the discussion of the trace (p. 38), for given i the set of the \sigma_i \sigma_j for all j is the same as G; hence application of any \sigma_i to the set of elements \alpha_1, \alpha_2, \dots, \alpha_r permutes its members. Then application of any \sigma_i to f(x) permutes its factors, and therefore does not change its coefficients. Since the only elements of E which are left fixed by all of the \sigma_i are those of F, f(x) is in F; and it is obviously separable. By (*) above, any polynomial in F having \alpha as a root has all of the \alpha_i as roots; consequently it is divisible by f(x). This means that f(x) is the minimal polynomial of \alpha. Thus E is a separable extension of F.

46 para 2/2-3     Let f_i(x) be the minimal polynomial of \omega_i.
46 para 4           [See the outline of the proof of Theorem 16, below.]
47/-4                 [Delete “other”.]
48 para 1/-2       of B which are the identity on F is equal
49/2                   \sigma G_B  \sigma^{-1} = G_B
49/3                   of F if and only if

________________________________

Outline of the proof of the Fundamental Theorem

The sections P1, P2, … below correspond to the eight paragraphs of the proof.

Abbreviations

\{ \cdots \}                 the set of \cdots
#\{ \cdots \}               the number of elements in \{ \cdots \}
E/F is normal   E is a normal extension of F
iso                      isomorphism
auto                    automorphism
isos of E/F        isomorphisms of E that leave F fixed
autos of E/F      automorphisms of E that leave F fixed
iff                        if and only if
\sigma |_B                     \sigma restricted to B

P1.  Part (1), proved here, establishes an iso of sets {subgroups of G} \leftrightarrow {subfields of E} under which G_B \leftrightarrow B. Here B = the fixed field of G_B, G_B = the group of autos of E/B.

P2.  The order of a finite group is the number of its elements; the index of a subgroup of a finite group is the number of its left cosets. Part (3), proved here, says: the order of G_B = (E/B) and the index of G_B = (B/F).

P3.  This paragraph and the next establish an iso of sets {left cosets of G_B} \leftrightarrow {isos of B/F} under which \sigma G_B \leftrightarrow \sigma |_B, where \sigma is any element of G. Note that \sigma |_B need not be an auto of B; it is an iso from B to some subfield of E, which may be B itself or another one. In this paragraph it is shown that: (i) the map from cosets to isos is well-defined—i.e., if \tau G_B = \sigma G_B then \tau |_B =  \sigma |_B; and (ii) the map is one-to-one. (i) is stated differently (“The elements of G in any one coset of G_B map B in the same way”); to prove it in the form just given, observe that \tau G_B = \sigma G_B implies \tau = \sigma \rho for some \rho in G_B, hence if b is in B, \tau (b) = \sigma \rho (b) = \sigma (b).

P4.  This paragraph shows that the map of P3 from cosets to isos is onto, hence is an iso of sets. Then by P2 (B/F) = #{left cosets of G_B} = #{isos of B/F}.

P5.  It is claimed that G_{\sigma (B)} = \sigma G_B \sigma^{-1}. Indeed, \tau is in G_{\sigma (B)} iff for all b in B, \tau \sigma (b) = \sigma (b), i.e. \sigma^{-1} \tau \sigma (b) = b, which means that \sigma^{-1} \tau \sigma is in G_B, i.e. \tau is in \sigma G_B \sigma^{-1}.

P6.  This paragraph proves an independent proposition, which we may state as:

Corollary 3 to Theorem 14.  E/F is normal iff (E/F) is finite and (E/F) = #{autos of E/F}.

P7.  B/F is normal
iff (B/F) = #{autos of B/F} (by P6 applied with E = B, since (B/F) is finite)
iff #{isos of B/F} = #{autos of B/F} (since by P4 (B/F) = #{isos of B/F})
iff {isos of B/F} = {autos of B/F} —i.e., for all \sigma, \sigma (B) = B
iff for all \sigma, G_{\sigma (B)} = G_B, i.e.  \sigma G_B \sigma^{-1} = G_B (by P5), i.e. \sigma G_B = G_B \sigma, i.e.
G_B is a normal subgroup of G.

P8.  If B/F and G_B are normal, the iso of P3 can be written G/G_B \leftrightarrow {autos of B/F}, and this is an iso of groups, since the multiplications correspond: \sigma G_B \tau G_B = (\sigma \tau) G_B \leftrightarrow (\sigma \tau)|_B = \sigma |_B \tau |_B. Thus part (2) is established.

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