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 $i$th equation, obtaining

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

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

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.