Unit 5 Algebra Notes PDF

# More about Roots ## Unit - Y ### Defintion If $f(x) = a_0x^n + a_1x^(n-1) + ... + a_(n-1)x + a_n$ in $F[x]$, then the derivative of $f(x)$, written as $f'(x)$ is the polynomial $f'(x) = na_0x^(n-1) + (n-1)a_1x^(n-2) + ... + (n-1)a_(n-1)x^(n-2) + ... + a_(n-1)$, in $F[x]$. ### Lemma For any $...

# More about Roots ## Unit - Y ### Defintion If $f(x) = a_0x^n + a_1x^(n-1) + ... + a_(n-1)x + a_n$ in $F[x]$, then the derivative of $f(x)$, written as $f'(x)$ is the polynomial $f'(x) = na_0x^(n-1) + (n-1)a_1x^(n-2) + ... + (n-1)a_(n-1)x^(n-2) + ... + a_(n-1)$, in $F[x]$. ### Lemma For any $f(x), g(x) \in F[x]$ and any $a \in F$: 1. $(f(x) + g(x))' = f'(x) + g'(x)$ 2. $(af(x))' = af'(x)$ 3. $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$ ### Proof Let $f(x) = a_0 + a_1x + ... + a_nx^m$ and $g(x) = b_0 + b_1x + ... + b_mx^n$ be any two polynomials in $F[x]$ and we assume that $m \ge n$. To prove (i): $(f(x) + g(x))' = (a_0 + a_1x + ... + a_nx^m) + (b_0 + b_1x + ... + b_mx^n)$ $= (a_0 + b_0) + (a_1 + b_1)x + ... + (a_n + b_n)x + ... + a_mx^m + b_nx^n$ $... + a_mx^(m-1) + ... + nb_nx^(n-1)$ $= (a_0 + b_0) + (a_1 + b_1)x + ... + (a_n + b_n)x + ... + a_mx^(m-1) + b_mx^(n-1)$ $= (a_1 + 2a_2x + ... + ma_mx^(m-1)) + (b_1 + 2b_2x + ... + nb_nx^(n-1))$ $= f'(x) + g'(x)$. i.e., $(f(x) + g(x))' = f'(x) + g'(x)$. To prove (ii): Let $d \in F$ Then $af(x) = da_0 + da_1x + da_2x^2 + ... + da_nx^n$ $(af(x))' = da_1 + 2da_2x + ... + nda_nx^(n-1)$ $= d(a_1 + 2a_2x + ... + na_nx^(n-1))$ $= df'(x)$. i.e., $(af(x))' = df'(x) \forall f(x) \in F[x], \forall d \in F$. To prove (iii): To prove (iii), from part (i) and (ii), it is enough to prove it in the highly special case $f(x) = x^i$ and $g(x) = x^j$ where Now, $f(x)g(x) = x^(i+j)$ $(f(x)g(x))' = (i + j)x^(i+j - 1)$ $= ix^(i+j - 1) + jx^(i+j - 1)$ $(f(x)g(x))' = ix^(i-1)x^j + jx^i(x^(j-1))$ Now, $f'(x)g(x) = (ix^(i-1))x^j$ and $f(x)g'(x) = x^i(jx^(j-1))$ :. $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$. ## Corollary If $f(x) \in F[x]$ is irreducible and if the characteristic of $F$ is 0, then $f(x)$ has no multiple roots. ### Proof: To prove. Let $f(x) \in F[x]$ be irreducible and $F$ be a field of characteristic 0. $f(x)$ has no multiple roots Suppose that $f(x)$ has a multiple root. By lemma, $f(x)$ and $f'(x)$ have a non-trivial common factor, say $p(x)$ But $f(x)$ is irreducible, so we have the factors of $f(x)$ are 1 or $f(x)$. :. Either $p(x) = 1$ or $p(x) = f(x)$. = $p(x)f(x) = f(x)$. Also since $p(x)$ is a factor of $f'(x)$, we can write $f'(x) = p(x)g(x)$ for some polynomial $g(x) \in F[x]$. $f'(x) = f(x)g(x)$ [: $f(x) = p(x)$] = a multiple of $f(x)$ = $f(x)f(x)$ But we know that deg $f(x) =$ deg $f'(x)$. :. $f(x)/ f'(x)$ is valid only when $f'(x) = 0$. Since $F$ is a field of characteristic zero and since $f'(x) = 0$, we have $f(x) = \text{constant}$. :. $f(x)$ has no roots, which is a contradiction to the fact that $f(x)$ has a multiple roots. :. $f(x)$ has no multiple roots. ## Lemma The polynomial $f(X) \in F[X]$ has a multiple root if and only if $f(x)$ and $f'(x)$ have a non-trivial (that is, of positive degrees common factor. ### Proof: If $f(x)$ has a multiple root $a$. Then $f(x) = (x - a)^mg(x)$, where $m \ge 1$. $f'(x) = [(x-a)^mg(x)]'$ = $m(x-a)^(m-1)g(x) + (x-a)^mg'(x)$ = $(x-a)^(m-1)(mg(x) + (x-a)g'(x))$ , since $m \ge 1$. $f'(x) = (x - a)^mr(x)$ => $f(x)$ and $f'(x)$ have the common factor $(x - a)$. converse If $f(x)$ and $f'(x)$ have a non-trivial common factor. ### To prove: $f(x)$ has a multiple root. Suppose $f(x)$ has no multiple root. Then $f(x) = (x - d_1)(x - d_2)...(x - d_n)$ where the $d_i$ are all distinct. Then $f'(x) = \frac{\sum_{i=1}^n \prod_{j\neq i} (x - d_j)}{x}$. Suppose $\prod_{j \neq i} (d_i - d_j) \neq 0$, since the roots are all distinct. If $f(x)$ and $f'(x)$ have a non-trivial common factor, they must have a common root, namely, any root of this common factor. But $f(x)$ and $f'(x)$ have no nontrivial common factor. Which is a contradiction to the hypothesis. :. $f(x)$ has a multiple root. ## Corollary If $F$ is a field of characteristic $p \neq 0$, then the polynomial $x^p - x \in F[x]$, for any $n$, has distinct roots. ### Proof: Let $F$ be a field of characteristic $p \neq 0$ and let $f(x) = x^p - x, \forall n$. To prove. $f(x)$ has distinct roots. Now, $ f(x) = x^p - x$ $f'(x) = px^(p-1) - 1$ = $0 - 1 = -1$ [: Since $F$ is of characteristic $p$]. :. $f(x) = x^p - 1$ and $f'(x) = -1$ :. $f(x)$ and $f'(x)$ has no non-trivial common factors. :. By above lemma, $f(x)$ has no multiple roots. :. All the roots of $f(x)$ are distinct. ## Definition: The extension $K$ of F is a simple extension of F if $K = F(a)$ for some $a$ in $K$. ## Theorem: If $F$ is of characteristic 0 and if $a, b$ are algebraic over $F$, then there exists an element $c \in F(a,b)$ such that $F(a, b) = F(c)$. ### Proof: Let $F$ be a field of characteristic 0 and $a,b$ be algebraic elements over $F$. To prove. There exists an element $c \in F(a,b)$ such that $F(a,b) = F(c)$. Let $f(x)$ and $g(x)$, of degrees $m$ and $n$, be the irreducible polynomials over $F$ satisfied by $a$ and $b$ respectively. i.e, $f(a) = 0$ and $g(b) = 0$ ($a,b$ - algebraic). Let $K$ be an extension of $F$ which contains all the roots of $f(x)$ and $g(x)$. Since $f(x) \in F[x]$ and since $F$ is a field of characteristic zero and by Corollary, we have 'f(x) has no multiple roots'. :. All the roots of $f(x)$ are distinct. Similarly, all the roots of the polynomial $g(x)$ are also distinct. Let the roots of $f(x)$ be $a = a_1, a_2, ... a_m$ and those of $g(x), b = b_1, b_2, ... b_n$. If $ j = 1 $ , then $b_j \neq b_1 = b$. Consider the equation, $a_i + \lambda b_j \neq a_i + \lambda b_1 = a + b $ We note that $\lambda = \frac{a_i - a}{b_j - b}$ is the only solution of the equation in $K$ for all $i$ and $j \neq 1$. Since $F$ is of characteristic 0, it has an infinite number of elements. :. It is always possible to choose an element $\lambda$ in $F$ such that $a_i + \lambda b_j \neq a_i + \lambda b_1$ for all $i$ and all $j \neq 1$. Let $c = a + \lambda b$. Then, we have to prove that $c \in F(a,b)$ and $F(c) = F(a,b)$. . Since $F(a, b)$ is the field which contains both $a$ and $b$ and also $\lambda \in F$, i.e., $\lambda$, $a$, $b$ are all in $F(a,b)$ => $a + \lambda b \in F(a,b)$ i.e., $c \in F(a,b)$. To prove $F(c) = F(a,b)$ Since $c \in F(a,b)$, we have $F(c) \subseteq F(a,b)$. It remains to prove that $F(a,b) \subseteq F(c)$. For this, we have to prove that both $a$ and $b$ are in $F(c)$. Since $b$ is a root of $g(x)$, we have $g(b) = 0$ over $F$ => $b$ satisfies the polynomial $g(x)$ over $F(c)$. Let $K = F(c)$. Then $b$ satisfies the polynomial $g(x)$ over $K$. :. $(x-b)$ is a factor of $g(x)$ over some extension of $K$. Moreover, if $h(x) = f(c - \lambda x)$ then $h(x) \in K[x]$. And $h(b) = f(c - \lambda b)$ = $f(a)$ = 0 [: $a$ is a root of $f(x)$] i.e, $x-b$ is a factor of the polynomial $h(x)$ over $K$. From (2) and (3), we see that both $g(x)$ and $h(x)$ have $(x-b)$ as a common factor over some extension of $K$. We have to prove that $(x-b)$ is the gcd of both $g(x)$ and $h(x)$ over some extension of $K$ For, if $b_j \neq b$ is another root of $g(x)$ $h(b_j) = f(c - \lambda b_j)$ $\neq 0$ [: By the choice of $\lambda$ , $c - \lambda b_j$ avoids all roots $a_i$ of $f(x)$]. i.e., $b_j \neq b$ is not a root of $h(x)$ for $j \neq 1$. i.e., Any factor of $g(x)$ other than $x - b$ is not a factor of $h(x)$ in some extension of $K$. Also, since $g(x)$ has no multiple roots, we have $(x - b_j) | g(x)$. From (2) and (4), we see that $(x-b)$ is the gcd of $g(x)$ and $h(x)$ in some extension of $K$. Then they have a non-trivial greatest common divisor over $K$ which must be a divisor of $x-b$. Since the degree of $x-b$ is 1, we see that the greatest common divisor of $g(x)$ and $h(x)$ in $K[x]$ is exactly $x-b$. Thus $x - b \in K[x]$ $b \in K$ i.e., $b \in F(c)$ [$K=F(c)$] Now, $c = a + \lambda b$ $a = c - \lambda b$ Since $b, c \in F(c)$, $\lambda \in F$, we obtain that $c - \lambda b \in F(c)$ i.e., $a \in F(c)$. :. Both $a, b \in F(c)$. But $a, b \in F(a,b)$ always :. $F(a,b) \subseteq F(c)$ From, and (5) $F(a, b) = F(c)$. ## Unit IV (Examples). (Inner product space) 2) In $F^2$ define for $u = (a_1, a_2)$ and $v = (b_1, b_2)$, $(u, v) = a_1\overline{b_1} + a_2\overline{b_2}$. It is easy to verify that this defines an inner product on $F^2$. ### Proof: Let $u, v, w \in F^2$ and $\alpha, \beta \in F$. Then, $u = (a_1, a_2)$, $v = (b_1, b_2)$ and $w = (c_1, c_2)$ where $a_1, a_2, b_1, b_2, c_1, c_2 \in F$. (i) $(u, v) = a_1\overline{b_1} + a_2\overline{b_2}$ $(v, u) = \overline{a_1}b_1 + \overline{a_2}b_2$ = $a_1\overline{b_1} + a_2\overline{b_2}$ => (1) (ii) $(\alpha u, v) = \alpha a_1\overline{b_1} + \alpha a_2\overline{b_2}$ = $\alpha(a_1\overline{b_1} + a_2\overline{b_2})$ = $\alpha(u, v)$ => (2) (iii) $(u, \alpha v) = a_1\overline{\alpha b_1} + a_2\overline{\alpha b_2}$ = $\alpha a_1\overline{b_1} + \alpha a_2\overline{b_2}$ = $\alpha(a_1\overline{b_1} + a_2\overline{b_2})$ = $\alpha(u, v)$ => (3) (iv) $(u, u) = a_1\overline{a_1} + a_2\overline{a_2}$ = $a_1\overline{a_1} + \overline{a_1}a_2 + \overline{a_2}a_1 + a_2\overline{a_2}$ = $2|a_1|^2 + 2Re(a_1\overline{a_2}) + 2|a_2|^2$ = $|a_1|^2 + |a_1|^2 + |a_2|^2 + 2Re(a_1\overline{a_2})$ = $|a_1|^2 + |a_1|^2 + |a_2|^2 + 2Re(a_2\overline{a_1})$ [: $a_1\overline{a_2} = \overline{a_2\overline{a_1}}$] = $|a_1|^2 + |a_1|^2 + |a_2|^2 + 2Re(a_2\overline{a_1})$ = $2|a_1|^2 + 2Re(a_2\overline{a_1}) + |a_2|^2 + |a_2|^2$ = $|a_1|^2 + |a_1|^2 + |a_2|^2 + |a_2|^2 + 2Re(a_2\overline{a_1})$ = $|a_1|^2 + |a_1|^2 + |a_2|^2 + 2Re(a_1\overline{a_2}) + |a_2|^2$ = $2|a_1|^2 + 2Re(a_1\overline{a_2}) + 2|a_2|^2$ = $2|a_1|^2 + 2Re(a_1\overline{a_2}) + 2|a_2|^2$ = $(u, u) \ge 0$. Also, $(u, u) = 0$ => $a_1\overline{a_1} + a_2\overline{a_2} + \overline{a_1}a_2 + \overline{a_2}a_1 = 0$ => $|a_1|^2 + |a_2|^2 = 0$ => $a_1 = 0$ and $a_2 = 0$ => $u = (0, 0)$ => $u = 0$ (v) $(\alpha u + \beta v, w) = (\alpha(a_1, a_2) + \beta(b_1, b_2), (c_1, c_2))$ = $(\alpha a_1 + \beta b_1, \alpha a_2 + \beta b_2), (c_1, c_2))$ = $(\alpha a_1 + \beta b_1)\overline{c_1} + (\alpha a_2 + \beta b_2)\overline{c_2}$ = $\alpha a_1\overline{c_1} + \alpha a_2\overline{c_2} + \beta b_1\overline{c_1} + \beta b_2\overline{c_2}$ = $\alpha(\alpha a_1\overline{c_1} + a_2\overline{c_2}) + \beta(vb_1\overline{c_1} + b_2\overline{c_2})$ = $\alpha(u, w) + \beta(v, w)$ From (2) and (5) $(\alpha u + \beta v, w) = \alpha(u, w) + \beta(v, w)$. $\forall u, v, w \in F^2$ and $\forall \alpha, \beta \in F$. :.$F^2$ is an inner product space. 3) Let $V$ be the set of all continuous complex-valued functions on the closed unit interval [0,1]. If $f(t), g(t) \in V$, define $(f(t), g(t)) = \int_0^1 f(t)\overline{g(t)} dt$. Verify that this defines an inner product space. ### Proof Let $f(t), g(t), h(t) \in V$ and $\alpha, \beta \in F$. (i) $(f(t), g(t)) = \int_0^1 f(t)\overline{g(t)} dt $ $(g(t), f(t)) = \int_0^1 g(t)\overline{f(t)} dt $ $(g(t), f(t)) = \int_0^1 \overline{f(t)}\overline{g(t)} dt $ = $\int_0^1 \overline{f(t)g(t)} dt$ = $\overline{\int_0^1 f(t)g(t) dt}$ = $\overline{(f(t), g(t))}$ :. $(f(t), g(t)) = \overline{(g(t), f(t))}$ (ii) $(\alpha f(t), g(t)) = \int_0^1 \alpha f(t)\overline{g(t)} dt$ = $\alpha \int_0^1 f(t)\overline{g(t)} dt$ = $\alpha (f(t), g(t))$ => (1) (iii) $(f(t), \alpha g(t)) = \int_0^1 f(t)\overline{\alpha g(t)} dt$ = $\int_0^1 f(t)\overline{\alpha} \overline{g(t)} dt$ = $\overline\alpha \int_0^1 f(t)\overline{g(t)} dt$ = $\overline\alpha (f(t), g(t))$ => (2) (iv) $(f(t), f(t)) = \int_0^1 f(t)\overline{f(t) dt}$ = $\int_0^1 |f(t)|^2 dt$ => $(f(t), f(t)) \ge 0$. Now, $(f(t), f(t)) = 0$ => $\int_0^1 |f(t)|^2 dt = 0 $ => $\int_0^1 |f(t)|^2 dt \ge 0$ => $|f(t)|^2 = 0$ => $f(t) = 0$ Thus $(f(t), f(t)) = 0$ <=> $f(t) = 0$. (v) $(\alpha f(t) + \beta g(t), h(t)) = \int_0^1 (\alpha f(t) + \beta g(t))\overline{h(t)} dt$ = $\int_0^1 \alpha f(t)\overline{h(t)} dt + \int_0^1 \beta g(t)\overline{h(t)} dt$ = $\alpha\int_0^1 f(t)\overline{h(t)} dt + \beta\int_0^1 g(t)\overline{h(t)} dt$ = $\alpha (f(t), h(t)) + \beta(g(t), h(t))$ => (3) $(f(t), h(t)) = \int_0^1 f(t)\overline{h(t)} dt$ $\alpha(f(t), h(t)) = \alpha \int_0^1 f(t)\overline{h(t)} dt$ $(g(t), h(t)) = \int_0^1 g(t)\overline{h(t)} dt$ $\beta(g(t), h(t)) = \beta\int_0^1 g(t)\overline{h(t)} dt$ :. $(\alpha f(t) + \beta g(t), h(t)) = \alpha (f(t), h(t)) + \beta(g(t), h(t))$. :. $V$ is an inner product space. ## Corollary: Any finite extension of a field of characteristic 0 is a simple extension. ### Proof: Proof is as above theorem. (Induction method). [It $a_1, a_2, ..., a_n$ are algebraic over $F$, then there is an element $c \in F(a_1, a_2, ..., a_n)$ such that $F(c) = F(a_1, a_2, ..., a_n)$. ## The elements of Galois Theory. ## Theorem: If $K$ is a field and if $\sigma_1, \sigma_2, ..., \sigma_n$ are distinct automorphisms of $K$, then it is impossible to find elements $a_1, a_2, .., a_n$, not all 0, in $K$ such that $a_1\sigma_1(u) + a_2\sigma_2(u) + ... + a_n\sigma_n(u) = 0$ for all $u \in K$. ### Proof: Assume that we could find $a_1, a_2, ..., a_n$, not all zero, in $K$ such that $a_1\sigma_1(u) + a_2\sigma_2(u) + ... + a_n\sigma_n(u) = 0$ for all $u \in K$. Then we could find such a relation having as few nonzero terms as possible; on renumbering we can assume that this minimal relation is $a_1\sigma_1(u) + a_2\sigma_2(u) + ... + a_m\sigma_m(u) = 0$ where $a_1, a_2, ..., a_m$ are all different from 0. case(i) $m = 1$ In this case, $u$ becomes, $a_1\sigma_1(u) = 0$ for all $u \in K$. $a_1 = 0$ This is a contradiction to our assumption. Thus we may assume that $m \ge 1$. Since the automorphisms are distinct, there is an element $c \in K$ such that $\sigma_1(c) \neq \sigma_m(c)$. Since $c \in K$ for all $u \in K$, relation (1) must also hold for $c u$. i.e., $a_1\sigma_1(cu) + a_2\sigma_2(cu) + ... + a_m\sigma_m(cu) = 0$ for all $u \in K$. Using the hypothesis that the $\sigma_i$s are automorphisms of $K$, this relation becomes $a_1\sigma_1(c)\sigma_1(u) + a_2\sigma_2(c)\sigma_2(u) + ...+ a_m\sigma_m(c)\sigma_m(u) = 0$ $a_1(\sigma_1(c) - \sigma_m(c))\sigma_1(u) + a_2(\sigma_2(c) - \sigma_m(c))\sigma_2(u) + ... + a_m(\sigma_m(c) - \sigma_m(c))\sigma_m(u) = 0$ = $a_2[(\sigma_2(c) - \sigma_m(c))\sigma_1(u) + ... + a_m(\sigma_m(c) - \sigma_m(c))\sigma_m(u) = 0$. If we put $b_i = a_i(\sigma_i(c) - \sigma_m(c))$ for $i = 2, 3, ..., m$ Then the $b_i$ are in $K$, $b_m = a_m(\sigma_m(c) - \sigma_1(c)) = 0$, since $a_m \neq 0$ and $\sigma_m(c) - \sigma_1(c) \neq 0$ yet $b_2 \sigma_2(u) + ... + b_m \sigma_m(u) = 0$ for all $u \in K$. This produces a shorter relation. This is a contradiction. :. Our assumption is wrong. :. It is impossible to find elements $a_1, a_2, ..., a_n$, not all 0, in $K$ such that $a_1 \sigma_1(u) + a_2\sigma_2(u) + ...+ a_n\sigma_n(u) = 0$ for all $u \in K$. ## Definition: (Fixed field) If $G$ is a group of automorphisms of $K$, then the fixed field of $G$ is the set of all elements $a \in K$ such that $\sigma(a) = a$ for all $\sigma \in G$. i.e., fixed field of $G = \{ a \in K | \sigma(a) = a, \forall \sigma \in G \}$ ## Lemma: The fixed field of $G$ is a subfield of $K$. ### Proof: Let $a,b$ be any two elements of the fixed field of $G$. Then $\sigma(a) = a$ and $\sigma(b)=b$ for all $a, b \in K$ and for all $\sigma \in G$. Now, $\sigma(a + b) = \sigma(a) + \sigma(b)$ [$ \sigma$ - automorphism ] = $a + b$ :. $a + b$ belongs to the fixed field of $G$. Similarly $\sigma(ab) = \sigma(a)\sigma(b)$ = $ab$ :. $ab$ belongs to the fixed field of $G$. Next, $b \in K$, $b' \in K$ $\sigma(b') = \sigma(b^n) = b^n$ :. $\sigma(b') = b'$ :. $b'$ belongs to the fixed field of $G$. :. The fixed field of $G$ is a subfield of $K$. ## Definition: Let $K$ be a field and let $F$ be a subfield of $K$. Then the group of automorphisms of $K$ relative to $F$, written $G(K,F)$, is the set of all automorphisms of $K$ leaving every element of $F$ fixed; that is, the automorphism $\sigma$ of $K$ is in $G(K,F)$ if and only if $\sigma(a) = a$ for every $a \in F$. ## Example: 1) Let $K$ be the field of complex numbers and let $F$ be the field of real numbers. Compute $G(K,F)$. If $\sigma$ is any automorphism of $K$. Since $i^2 = -1$, $(\sigma(i))^2 = (\sigma(i))^2 = (\sigma(-1)) = -1$. Here $\sigma(i) = \pm i$. In addition, $\sigma$ leaves every real number fixed, then for any $a + bi$ where $a,b$ are real, $\sigma(a + bi) = \sigma(a) + \sigma(b)\sigma(i)$ = $a + bi$. Each of these possibilities, namely the mapping $\sigma(a + bi) = a + bi$ and $\sigma(a + bi) = a - bi$ defines an automorphism of $K$, $\sigma$ being the identity automorphism and $\sigma$ complex-conjugation. Thus $G(K, F)$ is a group of order 2. 2) Let $F_0$ be the field of rational numbers and let $K = F_0 (\sqrt[3]{2}$, where $\sqrt[3]{2}$ is the real cube root of 8. Every element in $K$ is of the form $a_0 + a_1\sqrt[3]{2} + a_2(\sqrt[3]{2})^2$, where $a_0, a_1, a_2$ are rational numbers. If $\sigma$ is an automorphism of $K$, then $\sigma(\sqrt[3]{2}) = \sqrt[3]{\sigma(2)} = \sqrt[3]{2}$. Hence $\sigma(\sqrt[3]{2})^2 = \sigma(\sqrt[3]{2} * \sqrt[3]{2}) = \sigma(\sqrt[3]{2})\sigma(\sqrt[3]{2})$. However, there is only one real cube root of 8 lying in $K$. Therefore, there is only one value of $\sigma(\sqrt[3]{2})$ and since $K$ is a subfield of the real field, we must have that $\sigma(\sqrt[3]{2}) = \sqrt[3]{2}$ But then $\sigma(a_0 + a_1\sqrt[3]{2} + a_2(\sqrt[3]{2})^2) = a_0 + a_1\sqrt[3]{2} + a_2(\sqrt[3]{2})^2$. i.e., $\sigma$ is the identity automorphism of $K$. We thus see that $G(K, F_0)$ consists only of the identity map - and in this case the fixed field of $G(K, F_0)$ is not $F_0$ but is, in fact, larger, being all ## Lemma: $G(K,F)$ is a subgroup of the group of all automorphisms of $K$. ### Proof: We note that $G(K, F)$ is non-empty because atleast the identity automorphism $1$ of $K$ is in $G(K,F)$. i.e., $1(a) = a, \forall a \in F$ Now, let $g, f \in G(K, F)$, then $f(a) = a, \forall a \in F$ and $g(a) = a, \forall a \in