1. Binary operation on a set (S)

A binary operation () on a non-empty set (S) is a rule that assigns to every ordered pair ((a,b)\in S\times S) a unique element (ab\in S).

That is, [ *:S\times S\to S ] so the set (S) must be closed under the operation.

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2. Is ((\mathbb{Z}, \times)) a group?

No. The set of integers (\mathbb{Z}) with usual multiplication is not a group.

Justification:

• Closure: For (a,b\in\mathbb{Z}), (ab\in\mathbb{Z}). So closure holds.
• Associativity: ((ab)c=a(bc)) for all integers. So associativity holds.
• Identity: (1\in\mathbb{Z}) satisfies (a\cdot 1=1\cdot a=a). So identity exists.
• Inverse: For (a\in\mathbb{Z}), the multiplicative inverse would be (\frac{1}{a}). But (\frac{1}{a}\in\mathbb{Z}) only for (a=1) or (a=-1).
  For example, (2\in\mathbb{Z}), but (\frac12\notin\mathbb{Z}).

Hence the inverse axiom fails. Therefore ((\mathbb{Z}, \times)) is not a group.

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3. Is ((\mathbb{Q}^+, )) a group, where (ab=\frac{ab}{2})?

Yes, ((\mathbb{Q}^+, *)) is a group.

Let (a,b,c\in\mathbb{Q}^+).

Closure

[ a*b=\frac{ab}{2}\in\mathbb{Q}^+ ] since (\mathbb{Q}^+) is closed under multiplication and division by (2).

Associativity

[ (ab)c=\left(\frac{ab}{2}\right)c =\frac{\frac{ab}{2}\cdot c}{2} =\frac{abc}{4} ] Also, [ a(bc)=a\left(\frac{bc}{2}\right) =\frac{a\cdot \frac{bc}{2}}{2} =\frac{abc}{4} ] So, [ (ab)c=a(bc) ]

Identity element

Let (e\in\mathbb{Q}^+) be the identity. Then [ a*e=a ] [ \frac{ae}{2}=a ] [ e=2 ] Thus (2\in\mathbb{Q}^+) is the identity element.

Inverse element

For (a\in\mathbb{Q}^+), let its inverse be (b). Then [ a*b=2 ] [ \frac{ab}{2}=2 ] [ ab=4 ] [ b=\frac{4}{a} ] Since (a\in\mathbb{Q}^+), (\frac{4}{a}\in\mathbb{Q}^+).
So every element has an inverse.

Therefore, ((\mathbb{Q}^+, *)) is a group.

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4. Solve (3*x=6)

Using the operation, [ 3*x=\frac{3x}{2} ] So, [ \frac{3x}{2}=6 ] [ 3x=12 ] [ x=4 ]

Hence, [ \boxed{x=4} ]

1. Definitions

Permutation group:
Let (X) be a non-empty set. A bijection (f:X\to X) is called a permutation of (X). The set of all permutations of (X), denoted (S_X), forms a group under composition of mappings. Any subgroup of (S_X) is called a permutation group or a group of permutations.
If (X) has (n) elements, then (S_X) is denoted by (S_n), the symmetric group of degree (n).

Isomorphism between groups:
Let ((G,\cdot)) and ((H,*)) be two groups. A mapping
[ \phi:G\to H ] is called an isomorphism if:

1. (\phi) is bijective,
2. (\phi(ab)=\phi(a)*\phi(b)) for all (a,b\in G).

If such a mapping exists, then (G) and (H) are said to be isomorphic, written [ G\cong H. ]

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2. Proof that every group is isomorphic to a group of permutations

This is Cayley’s theorem.

Let (G) be any group. For each (a\in G), define a mapping [ L_a:G\to G ] by [ L_a(x)=ax,\qquad x\in G. ] This is called left multiplication by (a).

First, (L_a) is a permutation of (G).
If (L_a(x)=L_a(y)), then [ ax=ay. ] By left cancellation in (G), [ x=y. ] So (L_a) is injective.
Also, for any (y\in G), choose (x=a^{-1}y). Then [ L_a(x)=a(a^{-1}y)=y. ] So (L_a) is surjective. Hence (L_a) is a bijection, i.e. a permutation of (G). Therefore [ L_a\in S_G. ]

Now define [ \phi:G\to S_G ] by [ \phi(a)=L_a. ] For (a,b\in G), [ \phi(ab)(x)=L_{ab}(x)=(ab)x=a(bx)=L_a(L_b(x))=(L_a\circ L_b)(x). ] Thus [ \phi(ab)=\phi(a)\circ\phi(b). ] So (\phi) is a homomorphism.

If (\phi(a)=\phi(b)), then (L_a=L_b). In particular, taking (x=e), [ L_a(e)=L_b(e) ] so [ a e=b e, ] hence [ a=b. ] Therefore (\phi) is injective. Thus (\phi) is an isomorphism from (G) onto its image (\phi(G)).

But (\phi(G)={L_a:a\in G}) is a subgroup of (S_G), and hence a permutation group. Therefore [ G\cong \phi(G). ] So every group is isomorphic to a group of permutations.
If (|G|=n), then (G) is isomorphic to a subgroup of (S_n).

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2. Group isomorphic to ({1,w,w^2})

Let [ G={1,w,w^2}, ] where (w) is a primitive cube root of unity, so [ w^3=1,\qquad w\neq 1. ] This is a cyclic group of order (3).

By Cayley’s theorem, (G) is isomorphic to the subgroup of (S_3) given by its left regular representation. Label the elements as [ 1,\ w,\ w^2 ] and write them in positions (1,2,3), respectively.

Define [ \phi:G\to S_3 ] by [ \phi(a)=L_a,\qquad L_a(x)=ax. ]

Now compute:

[ L_1(x)=1\cdot x=x. ] So [ \phi(1)=\text{identity permutation}=(1)(2)(3). ]

[ L_w(1)=w,\qquad L_w(w)=w^2,\qquad L_w(w^2)=w^3=1. ] Thus [ \phi(w)=(1\ 2\ 3). ]

[ L_{w^2}(1)=w^2,\qquad L_{w^2}(w)=w^3=1,\qquad L_{w^2}(w^2)=w^4=w. ] Thus [ \phi(w^2)=(1\ 3\ 2). ]

Therefore [ G\cong A_3={ \operatorname{id},\ (1\ 2\ 3),\ (1\ 3\ 2)}\le S_3. ]

The correspondence is:

[ \begin{array}{c|c} G & A_3 \ \hline 1 & \operatorname{id}=(1)(2)(3)\ w & (1\ 2\ 3)\ w^2 & (1\ 3\ 2) \end{array} ]

This correspondence preserves multiplication. For example, [ w\cdot w=w^2 ] corresponds to [ (1\ 2\ 3)(1\ 2\ 3)=(1\ 3\ 2). ] Also, [ w\cdot w^2=1 ] corresponds to [ (1\ 2\ 3)(1\ 3\ 2)=\operatorname{id}. ] Hence ({1,w,w^2}) is isomorphic to the permutation group (A_3).

Definitions

Let $(G,*)$ and $(G',\circ)$ be groups.

A map

$$\phi:G\to G'$$

is called a homomorphism if

$$\phi(a*b)=\phi(a)\circ\phi(b)\qquad \forall a,b\in G.$$

The kernel of $\phi$ is

$$\ker\phi=\{a\in G:\phi(a)=e'\},$$

where $e'$ is the identity element of $G'$.

The image of $\phi$ is

$$\operatorname{Im}\phi=\phi(G)=\{\phi(a):a\in G\}\subseteq G'.$$

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Check whether $\phi:\mathbb C\to\mathbb C$, $\phi(x+iy)=y$, is a homomorphism

Take $\mathbb C$ as the additive group of complex numbers. Let

$$z_1=x_1+iy_1,\qquad z_2=x_2+iy_2.$$

Then

$$z_1+z_2=(x_1+x_2)+i(y_1+y_2).$$

So

$$\phi(z_1+z_2)=y_1+y_2=\phi(z_1)+\phi(z_2).$$

Hence $\phi$ is a homomorphism.

Kernel

$$\ker\phi=\{x+iy:\phi(x+iy)=0\} =\{x+iy:y=0\} =\{x:x\in\mathbb R\}.$$

Thus

$$\boxed{\ker\phi=\mathbb R}$$

i.e. the set of all real numbers.

Image

For any real number $r$, choose $z=0+ir$. Then

$$\phi(z)=r.$$

Therefore every real number is in the image, and no non-real number can be in the image. Hence

$$\boxed{\operatorname{Im}\phi=\mathbb R}$$

as a subset of $\mathbb C$.

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Prove: A group homomorphism is injective if and only if its kernel contains only the identity

Let $\phi:G\to G'$ be a group homomorphism. Let $e$ and $e'$ be the identities of $G$ and $G'$ respectively.

Part 1: If $\phi$ is injective, then $\ker\phi=\{e\}$

Suppose $\phi$ is injective. Let $a\in\ker\phi$. Then

$$\phi(a)=e'.$$

Also,

$$\phi(e)=e'.$$

So

$$\phi(a)=\phi(e).$$

Since $\phi$ is injective, we get

$$a=e.$$

Thus $\ker\phi\subseteq\{e\}$. But always $e\in\ker\phi$, because

$$\phi(e)=e'.$$

Hence

$$\ker\phi=\{e\}.$$

Part 2: If $\ker\phi=\{e\}$, then $\phi$ is injective

Suppose $\ker\phi=\{e\}$. Let $a,b\in G$ such that

$$\phi(a)=\phi(b).$$

Then

$$\phi(a)\circ \phi(b)^{-1}=e'.$$

Since $\phi$ is a homomorphism,

$$\phi(a)\circ \phi(b)^{-1}=\phi(a)\circ \phi(b^{-1}) =\phi(ab^{-1}).$$

Therefore

$$\phi(ab^{-1})=e'.$$

So

$$ab^{-1}\in\ker\phi=\{e\}.$$

Hence

$$ab^{-1}=e,$$

which gives

$$a=b.$$

Thus $\phi$ is injective.

Therefore,

$$\boxed{\phi\text{ is injective}\iff \ker\phi=\{e\}.}$$

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OR

Let $G$ be a group and $H$ a normal subgroup of $G$. Define

$$\sigma:G\to \frac GH,\qquad \sigma(a)=aH.$$

(i) Prove that $\frac GH$ is a group

The elements of $\frac GH$ are the left cosets of $H$ in $G$:

$$\frac GH=\{aH:a\in G\}.$$

Define the operation by

$$(aH)(bH)=abH.$$

Well-defined:
Suppose $aH=a'H$ and $bH=b'H$. Then

$$a'=ah_1,\qquad b'=bh_2$$

for some $h_1,h_2\in H$. Hence

$$a'b'=ah_1bh_2.$$

Since $H$ is normal, $b^{-1}h_1b\in H$. Therefore

$$a'b'=ab(b^{-1}h_1b)h_2\in abH.$$

So

$$a'b'H=abH.$$

Thus the operation is well-defined.

Closure:
For $aH,bH\in\frac GH$, we have

$$(aH)(bH)=abH\in\frac GH.$$

Associativity:

$$(aH\cdot bH)\cdot cH=(abH)(cH)=abcH,$$

and

$$aH\cdot(bH\cdot cH)=aH(bcH)=abcH.$$

So the operation is associative.

Identity:
The coset $H=eH$ is the identity, because

$$(aH)(H)=aH,\qquad (H)(aH)=aH.$$

Inverse:
For $aH\in\frac GH$, the element $a^{-1}H$ is its inverse, since

$$(aH)(a^{-1}H)=aa^{-1}H=H,$$

and

$$(a^{-1}H)(aH)=a^{-1}aH=H.$$

Hence $\frac GH$ is a group.

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(ii) Prove that $\sigma$ is a homomorphism onto $\frac GH$ whose kernel is $H$

For $a,b\in G$,

$$\sigma(ab)=abH=(aH)(bH)=\sigma(a)\sigma(b).$$

So $\sigma$ is a homomorphism.

For any coset $aH\in\frac GH$, we have

$$\sigma(a)=aH.$$

Thus every element of $\frac GH$ is the image of some element of $G$. Hence $\sigma$ is onto.

Now,

$$a\in\ker\sigma \iff \sigma(a)=H \iff aH=H \iff a\in H.$$

Therefore,

$$\boxed{\ker\sigma=H.}$$

Thus $\sigma$ is an onto homomorphism from $G$ to $\frac GH$ with kernel $H$.

Ring. A non-empty set $R$ together with two binary operations $+$ (addition) and $\cdot$ (multiplication) is called a ring if:

1. $(R, +)$ is an abelian group, i.e.
   • $a + (b + c) = (a + b) + c$ for all $a, b, c \in R$ (associativity),
   • there exists $0 \in R$ with $a + 0 = 0 + a = a$ (additive identity),
   • for each $a \in R$ there exists $-a \in R$ with $a + (-a) = 0$ (additive inverse),
   • $a + b = b + a$ for all $a, b \in R$ (commutativity).
2. Multiplication is associative: $a(bc) = (ab)c$ for all $a, b, c \in R$.
3. Distributive laws hold: $a(b + c) = ab + ac, \qquad (a + b)c = ac + bc \quad \text{for all } a, b, c \in R.$

If in addition $ab = ba$ for all $a, b \in R$, the ring is called a commutative ring; if there exists $1 \in R$, $1 \neq 0$, with $a\cdot 1 = 1 \cdot a = a$ for all $a$, it is called a ring with unity.

Field. A commutative ring with unity $F$ in which every non-zero element has a multiplicative inverse is called a field. That is, for every $a \in F$, $a \neq 0$, there exists $a^{-1} \in F$ such that $a\,a^{-1} = a^{-1}a = 1.$ Equivalently, in a field the non-zero elements form an abelian group under multiplication. Examples: $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ are fields; $\mathbb{Z}$ is not a field since $2 \in \mathbb{Z}$ has no inverse in $\mathbb{Z}$.

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To show: $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$ is a field.

Write $F = \mathbb{Q}(\sqrt{2})$. Clearly $F \subseteq \mathbb{R}$ and $F \neq \varnothing$ (take $a = b = 0$).

(i) Closure under addition. Let $a + b\sqrt{2},\; c + d\sqrt{2} \in F$. Then $(a + b\sqrt{2}) + (c + d\sqrt{2}) = (a + c) + (b + d)\sqrt{2}.$ Since $a + c,\; b + d \in \mathbb{Q}$, the sum belongs to $F$.

(ii) Associativity and commutativity of addition follow from the corresponding properties of $\mathbb{Q}$ (and $\mathbb{R}$), because addition is performed coefficient-wise.

(iii) Additive identity. $0 = 0 + 0\sqrt{2} \in F$, and $x + 0 = x$ for every $x \in F$.

(iv) Additive inverse. For $a + b\sqrt{2} \in F$, $-(a + b\sqrt{2}) = (-a) + (-b)\sqrt{2} \in F,$ since $-a, -b \in \mathbb{Q}$. Hence $(F, +)$ is an abelian group.

(v) Closure under multiplication. For $a + b\sqrt{2},\; c + d\sqrt{2} \in F$, $(a + b\sqrt{2})(c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + bd(\sqrt{2})^2 = (ac + 2bd) + (ad + bc)\sqrt{2}.$ Here $ac + 2bd \in \mathbb{Q}$ and $ad + bc \in \mathbb{Q}$, so the product lies in $F$.

(vi) Associativity and commutativity of multiplication hold, being inherited from $\mathbb{R}$; also the distributive laws hold in $F$ since they hold in $\mathbb{R}$.

(vii) Multiplicative identity. $1 = 1 + 0\sqrt{2} \in F$ and $1 \cdot x = x$ for all $x \in F$.

(viii) Multiplicative inverse. Let $a + b\sqrt{2} \in F$ with $a + b\sqrt{2} \neq 0$. Then not both $a$ and $b$ are zero, so $a^2 - 2b^2 \neq 0$ (if $a^2 - 2b^2 = 0$ then $\sqrt{2} = a/b \in \mathbb{Q}$, impossible). Now $\frac{1}{a + b\sqrt{2}} = \frac{a - b\sqrt{2}}{(a + b\sqrt{2})(a - b\sqrt{2})} = \frac{a - b\sqrt{2}}{a^2 - 2b^2} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2}\sqrt{2}.$ Since $\dfrac{a}{a^2 - 2b^2} \in \mathbb{Q}$ and $\dfrac{-b}{a^2 - 2b^2} \in \mathbb{Q}$, this inverse belongs to $F$. Also $(a + b\sqrt{2})\cdot\frac{a - b\sqrt{2}}{a^2 - 2b^2} = \frac{a^2 - 2b^2}{a^2 - 2b^2} = 1.$

Thus every non-zero element of $F$ has a multiplicative inverse in $F$.

Since $F$ is a commutative ring with unity in which every non-zero element is invertible, $\mathbb{Q}(\sqrt{2})$ is a field. $\blacksquare$

Theorem. Let (R,R') be rings and (\phi:R\to R') a ring homomorphism with kernel [ I=\ker\phi={a\in R:\phi(a)=0'}. ] Then:

i) (\phi) carries zero to zero, i.e. (\phi(0)=0').

Since (0=0+0) in (R), and (\phi) preserves addition, [ \phi(0)=\phi(0+0)=\phi(0)+\phi(0). ] Now add (-\phi(0)) on both sides in the additive group of (R'): [ \phi(0)-\phi(0)=\phi(0)+\phi(0)-\phi(0). ] Hence [ 0'=\phi(0). ] Therefore (\phi(0)=0').

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ii) (\phi(-a)=-\phi(a)) for every (a\in R).

Let (a\in R). Since (-a) is the additive inverse of (a), we have [ a+(-a)=0. ] Applying (\phi), [ \phi(a+(-a))=\phi(0). ] Because (\phi) preserves addition and by part (i), (\phi(0)=0'), we get [ \phi(a)+\phi(-a)=0'. ] Thus (\phi(-a)) is the additive inverse of (\phi(a)) in (R'). Therefore [ \phi(-a)=-\phi(a). ]

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iii) (I=\ker\phi) is an ideal of (R).

We prove that (I) is an additive subgroup of (R) and absorbs multiplication on both sides.

First, (I\neq\varnothing), since by part (i), [ \phi(0)=0', ] so (0\in I).

Now let (a,b\in I). Then (\phi(a)=0') and (\phi(b)=0'). Since (\phi) preserves addition and by part (ii), [ \phi(a-b)=\phi(a+(-b))=\phi(a)+\phi(-b) =\phi(a)-\phi(b)=0'-0'=0'. ] Hence (a-b\in I). Therefore (I) is an additive subgroup of (R).

Next, let (r\in R) and (a\in I). Then [ \phi(ra)=\phi(r)\phi(a)=\phi(r)0'=0', ] so (ra\in I). Similarly, [ \phi(ar)=\phi(a)\phi(r)=0'\phi(r)=0', ] so (ar\in I).

Thus (I) is closed under subtraction and under left and right multiplication by elements of (R). Hence (I) is a two-sided ideal of (R).

[ \boxed{I=\ker\phi \text{ is an ideal of } R.} ]

Answer either option. Model answers for both are given below.

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Option I

1. Why do we extend a field?

A field (F) may not contain roots of every polynomial in (F[x]). By extending (F) to a larger field (E), we create a field in which those missing roots exist. This is necessary for:

• solving polynomial equations,
• constructing splitting fields,
• building algebraic closures,
• studying finite fields and Galois theory.

Concrete example:
Consider (f(x)=x^2-2\in \mathbb{Q}[x]). It has no root in (\mathbb{Q}), because (\sqrt2) is irrational. But in the extension field

[ \mathbb{Q}(\sqrt2)={a+b\sqrt2:a,b\in\mathbb{Q}}, ]

the element (\alpha=\sqrt2) satisfies

[ \alpha^2-2=0. ]

Thus extending (\mathbb{Q}) to (\mathbb{Q}(\sqrt2)) gives a field containing the required root. Similarly, (x^2+1) has no root in (\mathbb{R}), but it has roots (\pm i) in the extension (\mathbb{C}).

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2. Kronecker’s theorem: existence of a root in an extension field

Theorem. Let (F) be a field and (f(x)\in F[x]) be a nonconstant polynomial. Then there exists an extension field (E) of (F) and an element (\alpha\in E) such that

[ f(\alpha)=0. ]

Proof.

Since (F[x]) is a unique factorization domain, (f(x)) has an irreducible factor. Let (p(x)\in F[x]) be an irreducible factor of (f(x)). It is enough to find a field extension of (F) containing a root of (p(x)), because then that element will also be a root of (f(x)).

Consider the principal ideal

[ (p(x))={p(x)q(x):q(x)\in F[x]} ]

in (F[x]). Since (F[x]) is a principal ideal domain and (p(x)) is irreducible, the ideal ((p(x))) is maximal. Therefore the quotient ring

[ E=F[x]/(p(x)) ]

is a field.

Let

[ \alpha=x+(p(x))\in E ]

be the coset of (x). We now show that (\alpha) is a root of (p(x)).

In the quotient field (E), the element (x) satisfies

[ p(x)+(p(x))=0+(p(x)). ]

Therefore

[ p(\alpha)=p(x)+(p(x))=0. ]

Hence (\alpha) is a root of (p(x)) in (E).

Now we must identify (F) as a subfield of (E). Define

[ \phi:F\to E,\qquad \phi(a)=a+(p(x)). ]

If (a\in F) and (\phi(a)=0), then (a\in (p(x))). But (p(x)) is nonconstant, so a nonzero constant polynomial (a) cannot be a multiple of (p(x)). Therefore (\phi(a)=0) implies (a=0). Thus (\phi) is injective. Hence (F) is isomorphic to the subfield (\phi(F)) of (E). Identifying (F) with (\phi(F)), we may regard (E) as an extension field of (F).

Since (p(x)\mid f(x)), we have (f(x)=p(x)q(x)) for some (q(x)\in F[x]). Therefore

[ f(\alpha)=p(\alpha)q(\alpha)=0\cdot q(\alpha)=0. ]

Thus there exists an extension field (E) of (F) containing an element (\alpha) such that (f(\alpha)=0). ∎

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OR

Option II

1. (\sqrt2) is algebraic and (\pi) is transcendental

Definitions.
Let (K/F) be a field extension. An element (\alpha\in K) is called algebraic over (F) if there exists a nonzero polynomial (g(x)\in F[x]) such that (g(\alpha)=0). If no such polynomial exists, then (\alpha) is called transcendental over (F).

For (\sqrt2):
We have

[ (\sqrt2)^2-2=0. ]

Thus (\sqrt2) is a root of the polynomial

[ x^2-2\in \mathbb{Q}[x]. ]

Therefore (\sqrt2) is algebraic over (\mathbb{Q}). Its minimal polynomial over (\mathbb{Q}) is (x^2-2), so

[ [\mathbb{Q}(\sqrt2):\mathbb{Q}]=2. ]

For (\pi):
The number (\pi) is transcendental over (\mathbb{Q}). This is a deep theorem proved by Lindemann in 1882. It follows from the Lindemann–Weierstrass theorem.

A quick explanation using that theorem is as follows. Suppose, for contradiction, that (\pi) is algebraic over (\mathbb{Q}). Then (i\pi) is also algebraic over (\mathbb{Q}). By the Lindemann–Weierstrass theorem, if (\alpha) is a nonzero algebraic number, then (e^\alpha) is transcendental. Taking (\alpha=i\pi), we get that (e^{i\pi}) is transcendental. But Euler’s identity gives

[ e^{i\pi}=-1, ]

which is algebraic over (\mathbb{Q}). This is a contradiction. Hence (\pi) cannot be algebraic over (\mathbb{Q}). Therefore (\pi) is transcendental over (\mathbb{Q}).

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2. Theorem: (\alpha) is algebraic over (F) iff ([F(\alpha):F]) is finite

Theorem. Let (K/F) be a field extension and (\alpha\in K). Then (\alpha) is algebraic over (F) if and only if

[ [F(\alpha):F]<\infty. ]

Proof.

(⇒) Suppose (\alpha) is algebraic over (F).

Then there exists a nonzero polynomial (g(x)\in F[x]) such that (g(\alpha)=0). Among all such polynomials, choose one of least degree. Let this minimal polynomial be

[ m(x)=a_0+a_1x+\cdots+a_nx^n,\qquad a_n\neq 0. ]

Its degree is (n). Since (m(x)) is irreducible over (F), the ideal ((m(x))) is maximal in (F[x]). Hence

[ F[x]/(m(x)) ]

is a field. The map

[ F[x]\to F(\alpha),\qquad h(x)\mapsto h(\alpha) ]

has kernel ((m(x))). Therefore

[ F[x]/(m(x))\cong F[\alpha]. ]

Since the quotient is a field, (F[\alpha]) is a field. But (F(\alpha)) is the smallest field containing (F) and (\alpha), so

[ F(\alpha)=F[\alpha]. ]

Now consider the set

[ {1,\alpha,\alpha^2,\ldots,\alpha^{n-1}}. ]

Spanning: Any element of (F(\alpha)) is of the form (h(\alpha)) for some (h(x)\in F[x]). By the division algorithm in (F[x]),

[ h(x)=q(x)m(x)+r(x), ]

where (\deg r(x)<n). Evaluating at (\alpha),

[ h(\alpha)=q(\alpha)m(\alpha)+r(\alpha)=r(\alpha). ]

Thus (h(\alpha)) is a linear combination of (1,\alpha,\ldots,\alpha^{n-1}) over (F).

Linear independence: Suppose

[ c_0+c_1\alpha+\cdots+c_{n-1}\alpha^{n-1}=0 ]

with (c_i\in F). Then the polynomial

[ r(x)=c_0+c_1x+\cdots+c_{n-1}x^{n-1} ]

has degree (<n) and satisfies (r(\alpha)=0). Since (m(x)) is the minimal polynomial of (\alpha), we must have (r(x)=0). Hence

[ c_0=c_1=\cdots=c_{n-1}=0. ]

Therefore ({1,\alpha,\ldots,\alpha^{n-1}}) is a basis of (F(\alpha)) over (F). Hence

[ [F(\alpha):F]=n<\infty. ]

(⇐) Suppose ([F(\alpha):F]=m<\infty).

Then (F(\alpha)) is an (m)-dimensional vector space over (F). Consider the (m+1) elements

[ 1,\alpha,\alpha^2,\ldots,\alpha^m. ]

Since they are more than the dimension of the space, they must be linearly dependent over (F). Hence there exist (c_0,c_1,\ldots,c_m\in F), not all zero, such that

[ c_0+c_1\alpha+\cdots+c_m\alpha^m=0. ]

Let

[ g(x)=c_0+c_1x+\cdots+c_mx^m\in F[x]. ]

Then (g(x)) is a nonzero polynomial and (g(\alpha)=0). Therefore (\alpha) is algebraic over (F).

Hence (\alpha) is algebraic over (F) if and only if ([F(\alpha):F]) is finite. ∎

Third Isomorphism Theorem

Let (G) be a group and let (H,K) be normal subgroups of (G) such that
[ K \leq H \leq G. ] We prove that [ \frac{G}{H} \cong \frac{G/K}{H/K}. ]

Since (K\trianglelefteq G), the quotient group (G/K) is well-defined. Also, because (K\leq H), the set [ H/K={hK:h\in H} ] is a subgroup of (G/K). Moreover, (H/K\trianglelefteq G/K), since for any (gK\in G/K) and (hK\in H/K), [ (gK)(hK)(gK)^{-1}=ghg^{-1}K\in H/K, ] because (H\trianglelefteq G).

Now define a map [ \varphi:G/K\to G/H ] by [ \varphi(gK)=gH. ]

1. (\varphi) is well-defined

Suppose (g_1K=g_2K). Then [ g_1^{-1}g_2\in K\leq H. ] Hence (g_1H=g_2H), so (\varphi(g_1K)=\varphi(g_2K)). Thus (\varphi) is well-defined.

2. (\varphi) is a homomorphism

For (g_1K,g_2K\in G/K), [ \varphi((g_1K)(g_2K)) =\varphi(g_1g_2K) =g_1g_2H =(g_1H)(g_2H) =\varphi(g_1K)\varphi(g_2K). ] So (\varphi) is a homomorphism.

3. (\varphi) is onto

For any (gH\in G/H), we have [ \varphi(gK)=gH. ] Hence (\varphi) is surjective.

4. Kernel of (\varphi)

We have [ gK\in \ker\varphi \iff \varphi(gK)=H \iff gH=H \iff g\in H \iff gK\in H/K. ] Therefore, [ \ker\varphi=H/K. ]

By the First Isomorphism Theorem, [ \frac{G/K}{\ker\varphi}\cong G/H, ] that is, [ \frac{G/K}{H/K}\cong \frac{G}{H}. ] Hence [ \boxed{\frac{G}{H}\cong \frac{G/K}{H/K}}. ]

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Verification in the group of integers

Take [ G=(\mathbb Z,+),\qquad H=2\mathbb Z,\qquad K=6\mathbb Z. ] Since (6\mathbb Z\leq 2\mathbb Z), we have (K\leq H). Also, every subgroup of an abelian group is normal, so (H,K\trianglelefteq G).

Now [ \frac{G}{H}=\frac{\mathbb Z}{2\mathbb Z}, ] and [ \frac{G}{K}=\frac{\mathbb Z}{6\mathbb Z}. ] Also, [ \frac{H}{K}=\frac{2\mathbb Z}{6\mathbb Z}. ] The subgroup (\frac{2\mathbb Z}{6\mathbb Z}) consists of the elements [ 0+6\mathbb Z,\quad 2+6\mathbb Z,\quad 4+6\mathbb Z. ] Thus the quotient [ \frac{G/K}{H/K}

\frac{\mathbb Z/6\mathbb Z}{2\mathbb Z/6\mathbb Z} ] has exactly two cosets, namely [ {0,2,4}+6\mathbb Z \quad\text{and}\quad {1,3,5}+6\mathbb Z. ] Hence it is isomorphic to (\mathbb Z/2\mathbb Z). Therefore, [ \frac{\mathbb Z/6\mathbb Z}{2\mathbb Z/6\mathbb Z} \cong \frac{\mathbb Z}{2\mathbb Z}. ] This agrees with [ \frac{G}{H}=\frac{\mathbb Z}{2\mathbb Z}. ] So the third isomorphism theorem is verified in this case.

Gaussian integer. A Gaussian integer is a complex number of the form
$z=a+bi,\qquad a,b\in\mathbb Z.$ The set of all Gaussian integers is denoted by
$\mathbb Z[i]=\{a+bi\mid a,b\in\mathbb Z\}.$

Norm. The norm of a Gaussian integer (z=a+bi) is
$N(z)=N(a+bi)=a^2+b^2=z\bar z,$ where (\bar z=a-bi). Thus (N(z)) is a nonnegative ordinary integer.

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Norm properties

1. (N(z)\ge 0), and
   $N(z)=0\iff z=0.$
2. Multiplicativity:
   $N(zw)=N(z)N(w)\qquad \forall z,w\in\mathbb Z[i].$
3. (N(z)=N(\bar z)=N(-z)=N(iz)).
4. (N(z)=1\iff z\in{\pm1,\pm i}), i.e. (z) is a unit in (\mathbb Z[i]).
5. If (z\mid w) in (\mathbb Z[i]), then (N(z)\mid N(w)) in (\mathbb Z).
6. Associates have the same norm: if (w=uz) with (u) a unit, then (N(w)=N(z)).
7. (N(z)) is the square of the usual modulus:
   $N(z)=|z|^2.$

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(\mathbb Z[i]) is an integral domain

A ring (R) is an integral domain if it is a commutative ring with unity and has no zero divisors.

First, (\mathbb Z[i]) is a subring of (\mathbb C). Indeed, for
$\alpha=a+bi,\qquad \beta=c+di\in\mathbb Z[i],$ we have
$\alpha+\beta=(a+c)+(b+d)i\in\mathbb Z[i],$ and
$\alpha\beta=(ac-bd)+(ad+bc)i\in\mathbb Z[i].$ Also (0,1\in\mathbb Z[i]), and (-\alpha\in\mathbb Z[i]). Hence (\mathbb Z[i]) is a subring of (\mathbb C). Since multiplication in (\mathbb C) is commutative, (\mathbb Z[i]) is commutative and has unity (1).

Now suppose
$\alpha\beta=0.$ Taking norms,
$N(\alpha\beta)=N(0)=0.$ But the norm is multiplicative, so
$N(\alpha\beta)=N(\alpha)N(\beta).$ Thus
$N(\alpha)N(\beta)=0.$ Since (N(\alpha)) and (N(\beta)) are nonnegative integers, one of them must be zero. But
$N(\alpha)=0\iff \alpha=0,$ and similarly for (\beta). Therefore
$\alpha=0\quad\text{or}\quad \beta=0.$ So (\mathbb Z[i]) has no zero divisors. Hence (\mathbb Z[i]) is an integral domain.

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Does (\mathbb Z[i]) form a Euclidean domain?

Yes. (\mathbb Z[i]) is a Euclidean domain with Euclidean function
$d(z)=N(z)=a^2+b^2.$

A Euclidean domain is an integral domain (R) with a function
$d:R\setminus\{0\}\to \mathbb Z_{\ge 0}$ such that for any (\alpha,\beta\in R) with (\beta\ne 0), there exist (q,r\in R) satisfying
$\alpha=\beta q+r,$ where either (r=0) or
$d(r)<d(\beta).$

We prove this for (\mathbb Z[i]).

Let
$\alpha,\beta\in\mathbb Z[i],\qquad \beta\ne 0.$ Consider the complex number
$\frac{\alpha}{\beta}=x+iy,\qquad x,y\in\mathbb Q.$ Choose integers (m,n) such that
$|x-m|\le \frac12,\qquad |y-n|\le \frac12.$ Put
$q=m+ni\in\mathbb Z[i].$ Then define
$r=\alpha-\beta q.$ If (r=0), the division algorithm is satisfied. If (r\ne 0), then

$$r=\beta\left(\frac{\alpha}{\beta}-q\right) =\beta\bigl((x-m)+i(y-n)\bigr).$$

Using multiplicativity of the norm,
$N(r)=N(\beta)\,N\left(\frac{\alpha}{\beta}-q\right).$ But

$$N\left(\frac{\alpha}{\beta}-q\right) =(x-m)^2+(y-n)^2 \le \frac14+\frac14=\frac12.$$

Hence
$N(r)\le \frac12 N(\beta)<N(\beta).$ Therefore
$d(r)=N(r)<N(\beta)=d(\beta).$

Thus the division algorithm holds in (\mathbb Z[i]) with the Euclidean function (d=N). Hence (\mathbb Z[i]) is a Euclidean domain.

Consequently, (\mathbb Z[i]) is also a principal ideal domain and a unique factorization domain.

Relation between roots and coefficients

Let the roots of
[ f(x)=x^4+p_1x^3+p_2x^2+p_3x+p_4=0 ] be (\alpha,\beta,\gamma,\delta). Then by Vieta’s formulas,

[ \alpha+\beta+\gamma+\delta=-p_1 ]

[ \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=p_2 ]

[ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-p_3 ]

[ \alpha\beta\gamma\delta=p_4 ]

So the coefficients determine the elementary symmetric functions of the roots.

Are these relations sufficient to solve the equation?

They determine the roots uniquely as a set, because the roots of a polynomial are uniquely determined by its coefficients. However, the relations alone are not sufficient as a practical method to find the roots explicitly. They give four nonlinear symmetric equations in four unknowns, and solving them directly is essentially equivalent to solving the original quartic. Additional methods such as factorization, Ferrari’s method, or numerical methods are needed to obtain the roots.

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Solve (2x^3+x^2-7x-6=0)

Divide by (2):

[ x^3+\frac12x^2-\frac72x-3=0 ]

Let the roots be (a,\ a+3,\ b), since the difference between two roots is (3).

By Vieta’s formulas for the cubic,

[ a+(a+3)+b=-\frac12 ]

[ 2a+b+3=-\frac12 ]

[ b=-2a-\frac72 ]

Also,

[ a(a+3)+b(a+a+3)=-\frac72 ]

Substitute (b=-2a-\frac72):

[ a^2+3a+\left(-2a-\frac72\right)(2a+3)=-\frac72 ]

[ a^2+3a-4a^2-13a-\frac{21}{2}=-\frac72 ]

[ -3a^2-10a-\frac{21}{2}=-\frac72 ]

[ -3a^2-10a-7=0 ]

[ 3a^2+10a+7=0 ]

[ (3a+7)(a+1)=0 ]

So,

[ a=-1 \quad \text{or} \quad a=-\frac73 ]

Now use the product of roots:

[ a(a+3)b=3 ]

For (a=-1),

[ b=-2(-1)-\frac72=-\frac32 ]

Product:

[ (-1)(2)\left(-\frac32\right)=3 ]

So this is valid.

For (a=-\frac73),

[ b=-2\left(-\frac73\right)-\frac72=\frac76 ]

Product:

[ \left(-\frac73\right)\left(\frac23\right)\left(\frac76\right) =-\frac{49}{27}\neq 3 ]

So this case is rejected.

Hence the roots are

[ \boxed{-\frac32,\ -1,\ 2} ]

Check: the roots (-1) and (2) differ by (3), as required.

Reciprocal equation. An equation that is unchanged when $x$ is replaced by $1/x$ (apart possibly from an overall sign) is called a reciprocal equation.

How they look. The coefficients equidistant from the two ends are equal (first kind) or equal in magnitude and opposite in sign (second kind):

$a_0x^n+a_1x^{n-1}+\dots+a_1x+a_0=0\quad\text{or}\quad a_0x^n+a_1x^{n-1}+\dots-a_1x-a_0=0 .$

• First kind, odd degree: $x=-1$ is a root.
• Second kind, odd degree: $x=1$ is a root.
• Second kind, even degree: both $x=1$ and $x=-1$ are roots.

Application. Recognising the type lets us remove the root $\pm1$ by factorisation. The remaining even-degree reciprocal equation of degree $2m$ is divided by $x^m$ and reduced, with the substitution $y=x+\dfrac1x$, to an equation of degree $m$. Thus a quintic is reduced to a quadratic.

Solve $2x^5-15x^4+37x^3-37x^2+15x-2=0$. The coefficients $2,-15,37,-37,15,-2$ are equal and opposite from the ends, and the degree is odd, so $x=1$ is a root. Synthetic division by $(x-1)$ gives

$(x-1)\left(2x^4-13x^3+24x^2-13x+2\right)=0 .$

Divide the quartic by $x^2$:

$2\left(x^2+\frac1{x^2}\right)-13\left(x+\frac1x\right)+24=0 .$

Put $y=x+\frac1x$, so $x^2+\frac1{x^2}=y^2-2$:

$2y^2-13y+20=0\;\Rightarrow\;(2y-5)(y-4)=0\;\Rightarrow\; y=\tfrac52\ \text{or}\ 4 .$

• $y=\tfrac52$: $2x^2-5x+2=0\Rightarrow x=2,\ \tfrac12$.
• $y=4$: $x^2-4x+1=0\Rightarrow x=2\pm\sqrt3$.

Roots: $x=1,\ 2,\ \tfrac12,\ 2+\sqrt3,\ 2-\sqrt3$.