Mathematics Core (MAT-401)

1. Binary operation and groups

(a) Definition of a binary operation

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 $a*b\in S$.
Thus, $*:S\times S\to S$ and the set $S$ must be closed under the operation.

(b) Is $(\mathbb Z,\times)$ a group?

No, the set of integers $\mathbb Z$ with usual multiplication is not a group.

Justification:

• Closure: For $a,b\in\mathbb Z$, $ab\in\mathbb Z$. True.
• Associativity: $(ab)c=a(bc)$ for all integers. True.
• Identity: $1\in\mathbb Z$ satisfies $a\cdot 1=1\cdot a=a$. True.
• Inverse: For $a\in\mathbb Z$, we need $b\in\mathbb Z$ such that [ a\cdot b=1. ] This is possible only for $a=1$ or $a=-1$. For example, $2$ has no integer inverse because $\frac12\notin\mathbb Z$. Also $0$ has no inverse.

Hence $(\mathbb Z,\times)$ is not a group.

(c) Is $(\mathbb Q^+,*)$ a group, where $a*b=\frac{ab}{2}$?

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

Let $a,b,c\in\mathbb Q^+$.

1. Closure:
   [ a*b=\frac{ab}{2}\in\mathbb Q^+ ] since $a,b>0$ and rational. So closure holds.

2. Associativity:
   [ (ab)c=\left(\frac{ab}{2}\right)c =\frac{\left(\frac{ab}{2}\right)c}{2} =\frac{abc}{4}. ] Also, [ a(bc)=a\left(\frac{bc}{2}\right) =\frac{a\left(\frac{bc}{2}\right)}{2} =\frac{abc}{4}. ] Hence $(a*b)*c=a*(b*c)$.

3. Identity element:
   Let $e$ be the identity. Then [ ae=a \implies \frac{ae}{2}=a \implies e=2. ] Also, [ 2a=\frac{2a}{2}=a. ] So the identity element is $e=2\in\mathbb Q^+$.

4. Inverse element:
   For $a\in\mathbb Q^+$, let $b$ be its inverse. Then [ a*b=2 \implies \frac{ab}{2}=2 \implies ab=4 \implies b=\frac4a. ] Since $a\in\mathbb Q^+$, $\frac4a\in\mathbb Q^+$. Thus every element has an inverse.

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

(d) 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} ]

Check: [ 3*4=\frac{3\cdot 4}{2}=6. ]

So the solution is $\boxed{x=4}$.

1. Permutation group

Let $A$ be a non-empty set. A permutation of $A$ is a bijection from $A$ onto itself. The set of all permutations of $A$, under composition of mappings, is the symmetric group $S_A$.

Any subgroup of $S_A$ is called a permutation group or a group of permutations.

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2. Isomorphism between groups

Let $(G,\ast)$ and $(H,\circ)$ be two groups. A mapping
$f:G\to H$ is called an isomorphism if

1. $f$ is a homomorphism, i.e.
   $f(a\ast b)=f(a)\circ f(b)\quad \text{for all } a,b\in G;$
2. $f$ is one-to-one and onto.

If such an isomorphism exists, we write
$G\cong H.$

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

Let $G$ be any group. For each element $a\in G$, define a mapping
$L_a:G\to G$ by
$L_a(x)=ax\quad \text{for all } x\in G.$

This is called left multiplication by $a$.

Step 1: $L_a$ is a permutation of $G$

If $L_a(x)=L_a(y)$, then
$ax=ay.$ By left cancellation in a group,
$x=y.$ So $L_a$ is one-to-one.

Also, for any $y\in G$, choose
$x=a^{-1}y.$ Then
$L_a(x)=a(a^{-1}y)=y.$ So $L_a$ is onto. Hence $L_a$ is a bijection, i.e. a permutation of $G$.

Thus
$L_a\in S_G.$

Step 2: Define a map from $G$ to $S_G$

Define
$\phi:G\to S_G$ by
$\phi(a)=L_a.$

Step 3: $\phi$ is a homomorphism

For $a,b,x\in G$, $(\phi(a)\circ \phi(b))(x)=L_a(L_b(x))=a(bx)=(ab)x=L_{ab}(x)=\phi(ab)(x).$ Therefore
$\phi(a)\circ \phi(b)=\phi(ab).$ So $\phi$ is a homomorphism.

Step 4: $\phi$ is one-to-one

Suppose
$\phi(a)=\phi(b).$ Then
$L_a=L_b.$ In particular, applying both sides to the identity element $e$, $L_a(e)=L_b(e),$ so
$ae=be,$ hence
$a=b.$ Thus $\phi$ is one-to-one.

Therefore $\phi$ is an isomorphism from $G$ onto its image $\phi(G)$. Since $\phi(G)$ is a subgroup of $S_G$, it is a group of permutations. Hence
$G\cong \phi(G).$

Thus every group is isomorphic to a group of permutations.

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4. A group isomorphic to $\{1,w,w^2\}$

Let
$w=e^{2\pi i/3}$ be a primitive cube root of unity. Then
$w^3=1,\qquad w\neq 1.$ The set
$U_3=\{1,w,w^2\}$ is a group under multiplication. It is cyclic of order $3$.

A group isomorphic to $U_3$ is the alternating group on three symbols: $A_3=\{I,\ (1\,2\,3),\ (1\,3\,2)\},$ under composition of permutations.

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5. Correspondence

The isomorphism is given by
$\phi:U_3\to A_3$ where
$\phi(1)=I,\qquad \phi(w)=(1\,2\,3),\qquad \phi(w^2)=(1\,3\,2).$

In table form:

This correspondence preserves multiplication. For example, $w\cdot w=w^2,$ and correspondingly, $(1\,2\,3)\circ(1\,2\,3)=(1\,3\,2).$ Also, $w^3=1,$ and $(1\,2\,3)^3=I.$

Hence
$\{1,w,w^2\}\cong A_3.$

Definition.
Let ((G,)) and ((G',\circ)) be groups with identities (e) and (e'). A map
[ \phi:G\to G' ] is called a homomorphism if
[ \phi(ab)=\phi(a)\circ\phi(b)\qquad \forall a,b\in G. ]

The kernel of (\phi) is
[ \ker\phi={a\in G:\phi(a)=e'}. ]

The image of (\phi) is
[ \operatorname{Im}\phi=\phi(G)={\phi(a):a\in 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 usual 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)=\phi\big((x_1+x_2)+i(y_1+y_2)\big)=y_1+y_2. ] Also, [ \phi(z_1)+\phi(z_2)=y_1+y_2. ] Hence [ \phi(z_1+z_2)=\phi(z_1)+\phi(z_2). ] Therefore (\phi) is a homomorphism.

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Kernel of (\phi)

[ \phi(x+iy)=0 \implies y=0. ] Thus [ \ker\phi={x+i0:x\in\mathbb R}=\mathbb R. ] So the kernel is the real axis.

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Image of (\phi)

For any (r\in\mathbb R), choose (z=0+ir). Then [ \phi(z)=\phi(0+ir)=r. ] So every real number is in the image. Hence [ \operatorname{Im}\phi={r+i0:r\in\mathbb R}=\mathbb R. ] Thus the image is also the real axis.

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Prove: A homomorphism is injective iff its kernel contains only the identity

Let (\phi:G\to G') be a group homomorphism with identity (e\in G) and (e'\in G').

((\Rightarrow)) Suppose (\phi) is injective.

Let (a\in\ker\phi). Then [ \phi(a)=e'. ] But also [ \phi(e)=e'. ] So [ \phi(a)=\phi(e). ] Since (\phi) is injective, we get [ a=e. ] Therefore, [ \ker\phi={e}. ]

((\Leftarrow)) Suppose (\ker\phi={e}).

Let (a,b\in G) and suppose [ \phi(a)=\phi(b). ] Then [ \phi(a)\phi(b)^{-1}=e'. ] Since (\phi) is a homomorphism, [ \phi(ab^{-1})=\phi(a)\phi(b)^{-1}=e'. ] Thus [ ab^{-1}\in\ker\phi={e}. ] So [ ab^{-1}=e \implies a=b. ] Hence (\phi) is injective.

Therefore, a homomorphism between groups is injective if and only if its kernel contains only the identity element.

Definition of a Ring

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

1. Closure under addition: $a + b \in R$ for all $a, b \in R$.
2. Associativity of addition: $(a + b) + c = a + (b + c)$ for all $a, b, c \in R$.
3. Existence of additive identity: There exists $0 \in R$ such that $a + 0 = 0 + a = a$ for all $a \in R$.
4. Existence of additive inverse: For each $a \in R$, there exists $-a \in R$ such that $a + (-a) = (-a) + a = 0$.
5. Commutativity of addition: $a + b = b + a$ for all $a, b \in R$.
6. Closure under multiplication: $a \cdot b \in R$ for all $a, b \in R$.
7. Associativity of multiplication: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ for all $a, b, c \in R$.
8. Distributive laws: $a \cdot (b + c) = a \cdot b + a \cdot c$ and $(a + b) \cdot c = a \cdot c + b \cdot c$ for all $a, b, c \in R$.

A ring with identity is one that also has a multiplicative identity $1 \neq 0$. A commutative ring is one in which $a \cdot b = b \cdot a$ for all $a, b \in R$.

Definition of a Field

A commutative ring with identity $F$ is called a field if every non-zero element has a multiplicative inverse in $F$. That is, for every $a \in F$ with $a \neq 0$, there exists $a^{-1} \in F$ such that

$a \cdot a^{-1} = a^{-1} \cdot a = 1.$

Equivalently, a field is a commutative division ring. Thus in a field $F$:

• $(F, +)$ is an abelian group,
• $(F \setminus \{0\}, \cdot)$ is an abelian group,
• Multiplication distributes over addition.

$\mathbb{Q}(\sqrt{2})$ is a Field

Let $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$. We show that it satisfies all field axioms under the usual addition and multiplication of real numbers.

1. Closure under addition

Let $x = a + b\sqrt{2}$ and $y = c + d\sqrt{2}$ be in $\mathbb{Q}(\sqrt{2})$, where $a, b, c, d \in \mathbb{Q}$. Then

$x + y = (a + c) + (b + d)\sqrt{2}.$

Since $a + c \in \mathbb{Q}$ and $b + d \in \mathbb{Q}$, we have $x + y \in \mathbb{Q}(\sqrt{2})$. So $\mathbb{Q}(\sqrt{2})$ is closed under addition.

2. Closure under multiplication

$x \cdot y = (a + b\sqrt{2})(c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + bd(\sqrt{2})^2$ $= (ac + 2bd) + (ad + bc)\sqrt{2}.$

Since $ac + 2bd \in \mathbb{Q}$ and $ad + bc \in \mathbb{Q}$, we have $x \cdot y \in \mathbb{Q}(\sqrt{2})$. So $\mathbb{Q}(\sqrt{2})$ is closed under multiplication.

3. Associativity and commutativity

Addition and multiplication of real numbers are associative and commutative. Since $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{R}$, these properties are inherited. Hence both operations are associative and commutative in $\mathbb{Q}(\sqrt{2})$.

4. Additive identity and inverse

The element $0 = 0 + 0\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ serves as the additive identity. For $x = a + b\sqrt{2}$, the element $-x = (-a) + (-b)\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ satisfies $x + (-x) = 0$. So additive inverses exist.

5. Multiplicative identity

The element $1 = 1 + 0\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ satisfies $x \cdot 1 = x$ for all $x \in \mathbb{Q}(\sqrt{2})$. So a multiplicative identity exists.

6. Multiplicative inverse

Let $x = a + b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ with $x \neq 0$. Then $a$ and $b$ are not both zero. Consider

$\frac{1}{x} = \frac{1}{a + b\sqrt{2}} = \frac{a - b\sqrt{2}}{(a + b\sqrt{2})(a - b\sqrt{2})} = \frac{a - b\sqrt{2}}{a^2 - 2b^2}.$

Since $\sqrt{2}$ is irrational, $a^2 - 2b^2 \neq 0$ whenever $a, b \in \mathbb{Q}$ are not both zero. Hence

$\frac{1}{x} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2}\sqrt{2} \in \mathbb{Q}(\sqrt{2}).$

Thus every non-zero element of $\mathbb{Q}(\sqrt{2})$ has a multiplicative inverse in $\mathbb{Q}(\sqrt{2})$.

7. Distributive laws

Since $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{R}$ and $\mathbb{R}$ is a field, the distributive laws hold automatically in $\mathbb{Q}(\sqrt{2})$.

Conclusion

All the field axioms are satisfied by $\mathbb{Q}(\sqrt{2})$ under the usual addition and multiplication. Therefore, $\mathbb{Q}(\sqrt{2})$ forms a field. $\blacksquare$

Let (0_R) and (0_{R'}) denote the zero elements of (R) and (R'), respectively.
Let (\phi:R\to R') be a ring homomorphism and let
[ I=\ker\phi={a\in R:\phi(a)=0_{R'}}. ]

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i) Prove that (\phi(0_R)=0_{R'})

Since (0_R+0_R=0_R), applying (\phi) gives [ \phi(0_R)=\phi(0_R+0_R)=\phi(0_R)+\phi(0_R). ] Now add (-\phi(0_R)) to both sides in (R'): [ \phi(0_R)-\phi(0_R)=\phi(0_R)+\phi(0_R)-\phi(0_R). ] Hence [ 0_{R'}=\phi(0_R). ] Therefore, [ \boxed{\phi(0_R)=0_{R'}}. ]

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ii) Prove that (\phi(-a)=-\phi(a)) for all (a\in R)

For any (a\in R), we have [ a+(-a)=0_R. ] Applying (\phi), [ \phi(a+(-a))=\phi(0_R). ] Since (\phi) is additive, [ \phi(a)+\phi(-a)=\phi(0_R). ] Using part (i), (\phi(0_R)=0_{R'}). Thus [ \phi(a)+\phi(-a)=0_{R'}. ] This means (\phi(-a)) is the additive inverse of (\phi(a)) in (R'). Therefore, [ \boxed{\phi(-a)=-\phi(a)\quad \forall a\in R}. ]

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

To prove (I) is an ideal of (R), we show:

1. (I\neq \varnothing).
2. (I) is closed under subtraction.
3. (I) absorbs multiplication from both sides.

1. Non-empty

From part (i), [ \phi(0_R)=0_{R'}. ] Hence (0_R\in I). Therefore (I\neq\varnothing).

2. Closed under subtraction

Let (a,b\in I). Then [ \phi(a)=0_{R'},\qquad \phi(b)=0_{R'}. ] Now [ \phi(a-b)=\phi(a+(-b))=\phi(a)+\phi(-b). ] By part (ii), (\phi(-b)=-\phi(b)). Hence [ \phi(a-b)=\phi(a)-\phi(b)=0_{R'}-0_{R'}=0_{R'}. ] Thus (a-b\in I). So (I) is an additive subgroup of (R).

3. Absorbs multiplication

Let (a\in I) and (r\in R). Then (\phi(a)=0_{R'}). Therefore, [ \phi(ra)=\phi(r)\phi(a)=\phi(r)0_{R'}=0_{R'}. ] Hence (ra\in I).

Similarly, [ \phi(ar)=\phi(a)\phi(r)=0_{R'}\phi(r)=0_{R'}. ] Hence (ar\in I).

Thus for every (r\in R) and (a\in I), [ ra\in I,\qquad ar\in I. ] So (I) absorbs multiplication from both sides.

Since (I) is non-empty, closed under subtraction, and closed under multiplication by elements of (R) on both sides, (I) is a two-sided ideal of (R).

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

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Hence all three statements are proved.

Answer (First alternative)

1. Why do we extend a field? Justify with a concrete example. [3]

A field extension (E/F) means a larger field (E) that contains (F) as a subfield. We extend a field mainly to:

• solve polynomial equations that have no root in the original field,
• adjoin roots of irreducible polynomials,
• construct algebraic closures,
• study symmetries of roots, which leads to Galois theory.

Concrete example:
In (\mathbb{Q}), the polynomial [ x^2-2 ] has no root, because (\sqrt2) is irrational. So we extend (\mathbb{Q}) to [ \mathbb{Q}(\sqrt2)={a+b\sqrt2 : a,b\in\mathbb{Q}}. ] This is a field, and in it the element (\alpha=\sqrt2) satisfies [ \alpha^2-2=0. ] Thus extending (\mathbb{Q}) to (\mathbb{Q}(\sqrt2)) gives a solution of (x^2-2=0). Also, [ [\mathbb{Q}(\sqrt2):\mathbb{Q}]=2. ]

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2. Kronecker’s theorem: existence of a root in an extension field. [7]

Theorem. Let (F) be a field and let (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, factor (f(x)) into irreducibles. Let (p(x)) be an irreducible factor of (f(x)) in (F[x]). Then (p(x)) is nonconstant and [ p(x)\mid f(x). ]

Consider the principal ideal [ I=(p(x)) ] in (F[x]). Because (F[x]) is a principal ideal domain and (p(x)) is irreducible, the ideal (I) is maximal. Hence the quotient ring [ E=F[x]/I ] is a field.

Let [ \alpha=x+I\in E. ] We now show that (E) is an extension field of (F). Define a map [ \varphi:F\to E,\qquad \varphi(a)=a+I. ] This is a ring homomorphism. Its kernel is [ \ker\varphi={a\in F: a+I=I}={a\in F: a\in (p(x))}. ] But (p(x)) is nonconstant, so it cannot divide a nonzero constant. Therefore [ \ker\varphi={0}. ] Thus (\varphi) is injective. Identifying (F) with its image (\varphi(F)), we may regard (E) as an extension field of (F).

Now evaluate (p(x)) at (\alpha): [ p(\alpha)=p(x+I)=p(x)+I=0+I=0_E. ] So (\alpha) is a root of (p(x)).

Since (p(x)\mid f(x)), write [ f(x)=p(x)q(x) ] for some (q(x)\in F[x]). Then [ f(\alpha)=p(\alpha)q(\alpha)=0\cdot q(\alpha)=0. ] Hence (\alpha\in E) is a root of (f(x)).

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

Note. This construction is called Kronecker’s construction. Also, [ [E:F]=\deg p(x)\le \deg f(x). ]

Answering the first alternative.

Prove (\displaystyle \frac{G}{H} \cong \frac{\frac{G}{K}}{\frac{H}{K}})

Let (G) be a group and let (H,K) be normal subgroups of (G) with (K\le H).

Since (K\trianglelefteq G), the quotient (G/K) is a group. Also, since (H\trianglelefteq G) and (K\le H), the set [ H/K={hK:h\in H} ] is a subgroup of (G/K).

Moreover, (H/K\trianglelefteq G/K). Indeed, 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) gives (ghg^{-1}\in H). Hence the quotient [ \frac{G/K}{H/K} ] is well-defined.

Define a map [ \phi:G/K\to G/H ] by [ \phi(gK)=gH. ]

1. (\phi) is well-defined

If (g_1K=g_2K), then (g_1^{-1}g_2\in K). Since (K\le H), we have (g_1^{-1}g_2\in H), so [ g_1H=g_2H. ] Thus (\phi(g_1K)=\phi(g_2K)).

2. (\phi) is a homomorphism

For (g_1K,g_2K\in G/K), [ \phi((g_1K)(g_2K)) =\phi(g_1g_2K) =g_1g_2H =(g_1H)(g_2H) =\phi(g_1K)\phi(g_2K). ]

3. (\phi) is onto

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

4. Kernel of (\phi)

Now [ gK\in \ker\phi \iff \phi(gK)=H \iff gH=H \iff g\in H \iff gK\in H/K. ] Therefore, [ \ker\phi=H/K. ]

By the First Isomorphism Theorem, [ \frac{G/K}{H/K}\cong G/H. ] Equivalently, [ \frac{G}{H}\cong \frac{\frac{G}{K}}{\frac{H}{K}}. ]

Hence the third isomorphism theorem is proved. (\square)

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Verification in the group of integers

Take [ G=\mathbb Z,\qquad K=6\mathbb Z,\qquad H=2\mathbb Z. ] Since (\mathbb Z) is abelian, every subgroup is normal. Also, [ 6\mathbb Z\le 2\mathbb Z, ] so (K\le H).

Now, [ G/H=\mathbb Z/2\mathbb Z\cong \mathbb Z_2. ] Also, [ G/K=\mathbb Z/6\mathbb Z\cong \mathbb Z_6. ] And [ H/K=2\mathbb Z/6\mathbb Z ={6\mathbb Z,;2+6\mathbb Z,;4+6\mathbb Z} \cong \mathbb Z_3. ] Therefore, [ \frac{G/K}{H/K} =\frac{\mathbb Z_6}{{0,2,4}}. ] The quotient has exactly two cosets: [ {0,2,4},\qquad {1,3,5}. ] Hence [ \frac{\mathbb Z_6}{{0,2,4}}\cong \mathbb Z_2. ] Thus [ \frac{G/K}{H/K}\cong \mathbb Z_2\cong G/H. ] So the result is verified for (G=\mathbb Z), (K=6\mathbb Z), and (H=2\mathbb Z).

8. Gaussian Integers and Euclidean Domain

1. Gaussian integer

A Gaussian integer is a complex number of the form
[ z=a+bi,\qquad a,b\in \mathbb Z,\quad i^2=-1. ] The set of all Gaussian integers is denoted by
[ \mathbb Z[i]={a+bi:a,b\in\mathbb Z}. ]

2. Norm of a Gaussian integer

The norm of (z=a+bi) is defined by
[ N(z)=z\bar z=(a+bi)(a-bi)=a^2+b^2. ] Thus (N(z)) is a non-negative integer.

3. Norm properties

For (z,w\in \mathbb Z[i]),

1. (N(z)\ge 0), and (N(z)=0\iff z=0).
2. (N(zw)=N(z)N(w)).
3. (N(z)=N(\bar z)).
4. (N(z)=z\bar z).
5. (N(z)=1\iff z\in{\pm1,\pm i}), i.e. (z) is a unit.
6. If (z\mid w), then (N(z)\mid N(w)).

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

Clearly (\mathbb Z[i]) is a commutative ring with unity (1=1+0i) under usual addition and multiplication.

To prove it has no zero divisors, let (z=a+bi,\ w=c+di\in\mathbb Z[i]). Then [ N(zw)=N(z)N(w). ] Indeed, [ \begin{aligned} N(zw) &=N((ac-bd)+(ad+bc)i)\ &=(ac-bd)^2+(ad+bc)^2\ &=a^2c^2+b^2d^2+a^2d^2+b^2c^2\ &=(a^2+b^2)(c^2+d^2)\ &=N(z)N(w). \end{aligned} ] Now suppose [ zw=0. ] Then [ N(zw)=N(0)=0. ] So [ N(z)N(w)=0. ] Since (N(z),N(w)\in\mathbb Z), either (N(z)=0) or (N(w)=0). Hence either (z=0) or (w=0). Therefore (\mathbb Z[i]) has no zero divisors.

Thus (\mathbb Z[i]) is a commutative ring with unity and no zero divisors, so it is an integral domain.

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

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

We must show the division algorithm: for any (\alpha,\beta\in\mathbb Z[i]) with (\beta\neq0), there exist (q,r\in\mathbb Z[i]) such that [ \alpha=\beta q+r, ] where either (r=0) or [ N(r)<N(\beta). ]

Let [ \frac{\alpha}{\beta}=x+yi,\qquad x,y\in\mathbb Q. ] Choose integers (m,n) nearest to (x,y), respectively, so that [ |x-m|\le \frac12,\qquad |y-n|\le \frac12. ] Set [ q=m+ni\in\mathbb Z[i]. ] Then define [ r=\alpha-\beta q. ] If (r=0), we are done. If (r\neq0), then [ r=\beta\left(\frac{\alpha}{\beta}-q\right) =\beta\bigl((x-m)+(y-n)i\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<1. ] Therefore [ N(r)<N(\beta). ] So the required division algorithm holds.

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

Relations 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 $f(x)=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$, and comparing coefficients:

$\alpha+\beta+\gamma+\delta=-p_1$ $\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=p_2$ $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-p_3$ $\alpha\beta\gamma\delta=p_4$

Are the relations sufficient to solve the equation? No, not by themselves. They are four symmetric equations in four unknowns, but they are symmetric in the roots, so eliminating three of the roots just gives back $f(x)=0$ for the fourth. Solving them is as hard as solving the original equation. They become useful only with extra information, such as a relation between the roots (two roots equal, in progression, with a given sum or difference).

Solve $2x^3+x^2-7x-6=0$, two roots differing by 3

Let the roots be $\alpha,\ \alpha+3,\ \beta$. Then

$2\alpha+3+\beta=-\tfrac12\;\Rightarrow\;\beta=-\tfrac72-2\alpha$ $\alpha(\alpha+3)+\beta(2\alpha+3)=-\tfrac72$ $\alpha(\alpha+3)\beta=3$

Substituting $\beta$ in the second relation:

$\alpha^2+3\alpha+\left(-\tfrac72-2\alpha\right)(2\alpha+3)=-\tfrac72$ $-3\alpha^2-10\alpha-7=0\;\Rightarrow\;3\alpha^2+10\alpha+7=0\;\Rightarrow\;(3\alpha+7)(\alpha+1)=0 .$

So $\alpha=-1$ or $\alpha=-\tfrac73$.

• $\alpha=-1$: $\beta=-\tfrac72+2=-\tfrac32$, roots $-1,\ 2,\ -\tfrac32$. Product $=(-1)(2)\left(-\tfrac32\right)=3$ ✓.
• $\alpha=-\tfrac73$: roots $-\tfrac73,\ \tfrac23,\ \tfrac76$; the product is $-\tfrac{49}{27}\ne3$, so this is rejected.

Check: $2x^3+x^2-7x-6=(x-2)(x+1)(2x+3)$.

$\boxed{x=2,\quad x=-1,\quad x=-\tfrac32}$

Part A — Reciprocal equations and the solution of $2x^5-15x^4+37x^3-37x^2+15x-2=0$

Definition

A reciprocal equation is an equation which is unchanged, or changes only in sign, when the variable $x$ is replaced by its reciprocal $\dfrac1x$. In symbols, $f(x)=0$ is reciprocal if

$x^{\,n}f\!\left(\tfrac1x\right)=\pm\,f(x).$

How they look

Written in the standard descending form

$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=0,$

the coefficients read the same from either end:

• First kind: $a_r=a_{n-r}$ for every $r$ — e.g. $x^4-5x^3+6x^2-5x+1=0$.
• Second kind: $a_r=-a_{n-r}$ for every $r$ — e.g. $2x^5-15x^4+37x^3-37x^2+15x-2=0$.

Application of recognising a reciprocal equation

1. Roots occur in reciprocal pairs: if $\alpha$ is a root, so is $\dfrac1\alpha$.
2. Odd degree is reduced at once: if $a_0=-a_n$, then $x=1$ is a root; if $a_0=a_n$, then $x=-1$ is a root. Dividing by $(x\mp1)$ leaves an even-degree reciprocal equation.
3. Even degree is halved: dividing the even-degree equation by $x^{n/2}$ and putting $y=x+\dfrac1x$ (so that $x^2+\dfrac1{x^2}=y^2-2$, etc.) lowers the degree to $n/2$, which is then solved as an ordinary quadratic.
4. Hence the method gives a quick, systematic solution of palindromic polynomial equations instead of trial-and-error root hunting.

Solution

$2x^5-15x^4+37x^3-37x^2+15x-2=0 \qquad \cdots(1)$

Here $a_0=2$ and $a_5=-2$, so $a_0=-a_5$: it is a reciprocal equation of odd degree of the second kind. Hence $x=1$ is a root. Dividing (1) by $(x-1)$ (synthetic division with coefficients $2,-15,37,-37,15,-2$):

$2,\;-13,\;24,\;-13,\;2,\;0$

so that

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

The quartic is an even-degree reciprocal equation. Divide it by $x^2$:

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

Put $y=x+\dfrac1x$, so that $x^2+\dfrac1{x^2}=y^2-2$:

$2(y^2-2)-13y+24=0$ $2y^2-13y+20=0$ $(2y-5)(y-4)=0 \;\Rightarrow\; y=4,\; y=\tfrac52.$

Case 1: $y=4$

$x+\frac1x=4 \;\Rightarrow\; x^2-4x+1=0 \;\Rightarrow\; x=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3 .$

Case 2: $y=\dfrac52$

$x+\frac1x=\frac52 \;\Rightarrow\; 2x^2-5x+2=0 \;\Rightarrow\; (2x-1)(x-2)=0 \;\Rightarrow\; x=\frac12,\; x=2 .$

Answer:

$\boxed{x=1,\quad x=2+\sqrt3,\quad x=2-\sqrt3,\quad x=2,\quad x=\frac12}$

(Note the reciprocal pairs: $2+\sqrt3$ with $2-\sqrt3$, and $2$ with $\tfrac12$.)

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Part B (OR) — Cardan's method and the solution of $9x^3-6x^2+1=0$

Cardan's method

Let the cubic be

$x^3+ax^2+bx+c=0. \qquad \cdots(1)$

Step 1 — Remove the $x^2$ term. Put $x=y-\dfrac a3$. Then (1) becomes the depressed cubic

$y^3+py+q=0,$

where

$p=b-\frac{a^2}{3},\qquad q=\frac{2a^3}{27}-\frac{ab}{3}+c .$

Step 2 — Split the unknown. Put $y=u+v$. Then

$y^3=u^3+v^3+3uv(u+v)=u^3+v^3+3uv\,y .$

Choose $u,v$ so that $3uv=-p$, i.e. $uv=-\dfrac p3$. Then

$y^3=u^3+v^3-py \;\Rightarrow\; y^3+py-\left(u^3+v^3\right)=0 .$

Comparing with $y^3+py+q=0$ gives

$u^3+v^3=-q,\qquad u^3v^3=-\frac{p^3}{27}.$

Step 3 — Solve for $u^3,v^3$. They are the roots of the resolvent quadratic

$t^2+qt-\frac{p^3}{27}=0,$

so that

$u^3=-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}},\qquad v^3=-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} .$

Step 4 — Back-substitute. $y=u+v$, hence $x=y-\dfrac a3$. The remaining two roots are obtained from the quadratic factor left after removing the known root.

Nature of the roots (from the discriminant $\Delta=\dfrac{q^2}{4}+\dfrac{p^3}{27}$):

• $\Delta>0$: one real root and two complex conjugate roots;
• $\Delta=0$: all roots real, at least two equal;
• $\Delta<0$: three distinct real roots (the casus irreducibilis, handled by trigonometry).

Solution

$9x^3-6x^2+1=0 \;\Rightarrow\; x^3-\frac23x^2+\frac19=0 .$

Comparing with $x^3+ax^2+bx+c=0$: $\;a=-\dfrac23,\; b=0,\; c=\dfrac19$.

Step 1: Put $x=y-\dfrac a3=y+\dfrac29$. Then

$p=b-\frac{a^2}{3}=0-\frac{4/9}{3}=-\frac4{27},$

$q=\frac{2a^3}{27}-\frac{ab}{3}+c=\frac{2\left(-\frac{8}{27}\right)}{27}-0+\frac19=-\frac{16}{729}+\frac{81}{729}=\frac{65}{729}.$

So the depressed cubic is

$y^3-\frac4{27}y+\frac{65}{729}=0 .$

Step 2: Put $y=u+v$ with $uv=\dfrac4{81}$ and $u^3+v^3=-\dfrac{65}{729}$.

Step 3: $u^3,v^3$ are the roots of

$t^2+\frac{65}{729}t+\frac{64}{531441}=0 .$

$$t=\frac{-\dfrac{65}{729}\pm\sqrt{\dfrac{4225}{531441}-\dfrac{256}{531441}}}{2} =\frac{-\dfrac{65}{729}\pm\dfrac{63}{729}}{2}.$$

Hence

$u^3=-\frac1{729}\;\Rightarrow\; u=-\frac19,\qquad v^3=-\frac{64}{729}\;\Rightarrow\; v=-\frac49 .$

(Check: $uv=\dfrac4{81}$ ✓)

Step 4:

$y=u+v=-\frac19-\frac49=-\frac59,$

$x=y+\frac29=-\frac59+\frac29=-\frac13 .$

The remaining roots come from the quadratic factor. Dividing $9x^3-6x^2+1$ by $\left(x+\dfrac13\right)$:

$9x^3-6x^2+1=(3x+1)(3x^2-3x+1).$

$3x^2-3x+1=0 \;\Rightarrow\; x=\frac{3\pm\sqrt{9-12}}{6}=\frac{3\pm i\sqrt3}{6}=\frac12\pm\frac{i\sqrt3}{6}.$

Answer:

$\boxed{x=-\frac13,\qquad x=\frac{3+i\sqrt3}{6},\qquad x=\frac{3-i\sqrt3}{6}}$

Since $\Delta=\dfrac{q^2}{4}+\dfrac{p^3}{27}=\dfrac{3969}{2125764}>0$, the cubic has one real root $\left(-\tfrac13\right)$ and two complex conjugate roots, as obtained.