Section-A
(Very Short Answer Type Questions)
Note:- Answer all questions. As per the nature of the question you delimit your answer in one word, one sentence or maximum up to 30 words. Each question carries 1 (one) mark.
(i). If H and K are two sub groups of G such that G is an internal direct product of H and K . Then find H nn KH \cap K.
(ii). Write class equation for the finite group.
(iii). Define normal extension of a field.
(iv). Write Schwartz inequality.
Section – B
(Short Answer Questions)
Note :- Answer any two questions . Each answer should be given in 200 words. Each question carries 4 marks.
Prove that every Euclidean ring is a principal ideal domain.
If W is any subspace of a finite dimensional inner product space V , the prove that (W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W.
Prove that every additive abelian group is a module over the ring Z of integers.
If KK is a field and d_(1),d_(2),dots.d_(n)d_1, d_2, \ldots . d_n are distinct automorphisms of KK. Then prove that it is impossible to find elements b_(1),b_(2),dots.b_(n)b_1, b_2, \ldots . b_n not all zero in KK such that b_(1)phi_(1)(a)+b_(2)phi_(2)(a)+dots dots dots+b_(n)phi _(n)(a)=0AA a in Kb_1 \phi_1(a)+b_2 \phi_2(a)+\ldots \ldots \ldots+b_n \phi_n(a)=0 \forall a \in K.
Section – C
(Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
Let M_(1)M_1 and M_(2)M_2 are submodule of an R-module M . Then prove that:
Let V and V^(‘)V^{\prime} be any two finite dimensional vector space over the same field F . Then prove that the vector space Hom(V,V^(‘))\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right) of all linear transformation of V to V^(‘)V^{\prime} is also finite dimensional and dim Hom(V,V^(‘))=dimVxx dimV^(‘)\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right)=\operatorname{dim} \mathrm{V} \times \operatorname{dim} V^{\prime}.
Answer:
Question:-01(a)
If H and K are two subgroups of G such that G is an internal direct product of H and K. Then find H nn KH \cap K.
Answer:
If GG is the internal direct product of the subgroups HH and KK, the following conditions must hold:
Every element of GG can be uniquely written as a product hkh k where h in Hh \in H and k in Kk \in K.
H nn K={e}H \cap K = \{ e \}, where ee is the identity element of GG.
Thus, H nn K={e}H \cap K = \{ e \}.
Question:-01(b)
Write class equation for the finite group.
Answer:
The class equation of a finite group GG is a fundamental result in group theory that relates the structure of GG to its conjugacy classes. It is expressed as:
Z(G)Z(G): The center of GG, i.e., the set of elements that commute with every element of GG,
|Z(G)||Z(G)|: The size (order) of the center Z(G)Z(G),
g_(i)g_i: A representative of each conjugacy class outside the center,
C_(G)(g_(i))C_G(g_i): The centralizer of g_(i)g_i in GG, i.e., the set of elements in GG that commute with g_(i)g_i,
[G:C_(G)(g_(i))][G : C_G(g_i)]: The index of the centralizer C_(G)(g_(i))C_G(g_i) in GG, which equals the size of the conjugacy class containing g_(i)g_i.
Interpretation:
The elements in Z(G)Z(G) form singleton conjugacy classes (since they commute with all elements).
The remaining conjugacy classes contribute terms of the form [G:C_(G)(g_(i))][G : C_G(g_i)], representing their sizes.
The class equation is a powerful tool to analyze the structure of GG, especially in applications like Sylow theorems and proving group properties.
Question:-01(c)
Define normal extension of a field.
Answer:
A normal extension of a field is a type of field extension that satisfies a specific algebraic condition related to the splitting of polynomials. Formally:
Let EE be a field extension of FF (F sube EF \subseteq E). The extension E//FE/F is called a normal extension if:
Every irreducible polynomial f(x)in F[x]f(x) \in F[x] that has at least one root in EEsplits completely into linear factors over EE, i.e., all the roots of f(x)f(x) are contained in EE.
Equivalent Definitions:
E//FE/F is normal if EE is the splitting field of some family of polynomials in F[x]F[x].
E//FE/F is normal if it is closed under conjugation by automorphisms, i.e., for any field automorphism sigma\sigma of a larger field containing EE, if sigma(e)in E\sigma(e) \in E for some e in Ee \in E, then sigma(e)\sigma(e) is also in EE.
Examples:
The extension Q(sqrt2)//Q\mathbb{Q}(\sqrt{2})/\mathbb{Q} is not normal, because the irreducible polynomial x^(2)-2x^2 – 2 over Q\mathbb{Q} has a root sqrt2\sqrt{2} in Q(sqrt2)\mathbb{Q}(\sqrt{2}), but its other root, -sqrt2-\sqrt{2}, is not in Q(sqrt2)\mathbb{Q}(\sqrt{2}).
The extension Q(sqrt2,root(3)(3))//Q\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})/\mathbb{Q} is normal, because it is the splitting field of (x^(2)-2)(x^(3)-3)(x^2 – 2)(x^3 – 3) over Q\mathbb{Q}.
Normal extensions are fundamental in the study of Galois theory, where they play a key role in defining Galois extensions.
Question:-01(d)
Write Schwartz inequality.
Answer:
The Schwarz inequality (or Cauchy-Schwarz inequality) is a fundamental result in mathematics, particularly in linear algebra and analysis. It states:
For any two vectors u,v\mathbf{u}, \mathbf{v} in an inner product space, the following inequality holds:
(:u,v:)\langle \mathbf{u}, \mathbf{v} \rangle: The inner product of u\mathbf{u} and v\mathbf{v},
||u||=sqrt((:u,u:))\|\mathbf{u}\| = \sqrt{\langle \mathbf{u}, \mathbf{u} \rangle}: The norm (magnitude) of u\mathbf{u},
||v||=sqrt((:v,v:))\|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}: The norm (magnitude) of v\mathbf{v}.
Equality Condition:
Equality holds if and only if u\mathbf{u} and v\mathbf{v} are linearly dependent, i.e., there exists a scalar lambda\lambda such that u=lambdav\mathbf{u} = \lambda \mathbf{v}.
Special Cases:
In Euclidean space R^(n)\mathbb{R}^n, with the standard dot product:|u*v| <= ||u||||v||,|\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\|,where ||u||=sqrt(sum_(i=1)^(n)u_(i)^(2))\|\mathbf{u}\| = \sqrt{\sum_{i=1}^n u_i^2}.
For functions f,gf, g in an inner product space L^(2)[a,b]L^2[a, b]:|int_(a)^(b)f(x)g(x)dx| <= sqrt(int_(a)^(b)f(x)^(2)dx)*sqrt(int_(a)^(b)g(x)^(2)dx).\left| \int_a^b f(x)g(x) \, dx \right| \leq \sqrt{\int_a^b f(x)^2 \, dx} \cdot \sqrt{\int_a^b g(x)^2 \, dx}.
The Schwarz inequality is widely used in diverse areas of mathematics, physics, and engineering to establish bounds and relationships between quantities.
Question:-02
Prove that every Euclidean ring is a principal ideal domain.
Answer:
Statement:
Every Euclidean ring is a principal ideal domain (PID).
Proof:
Definitions:
Euclidean Ring: A commutative ring RR with unity is a Euclidean ring if there exists a function delta:R\\{0}rarrN\delta: R \setminus \{0\} \to \mathbb{N} (called a Euclidean norm) such that for all a,b in Ra, b \in R with b!=0b \neq 0:
There exist q,r in Rq, r \in R such that a=bq+ra = bq + r, where either r=0r = 0 or delta(r) < delta(b)\delta(r) < \delta(b).
Principal Ideal Domain (PID): A commutative ring RR with unity is a PID if every ideal I sube RI \subseteq R is a principal ideal, i.e., there exists some a in Ra \in R such that I=(a)={ra:r in R}I = (a) = \{ ra : r \in R \}.
Proof:
Let RR be a Euclidean ring, and let II be a nonzero ideal of RR. We need to show that II is a principal ideal.
Since II is nonzero, there exists some b in Ib \in I with b!=0b \neq 0. Among all nonzero elements of II, choose an element d in Id \in I such that delta(d)\delta(d) is minimal (this is possible because delta\delta maps elements of R\\{0}R \setminus \{0\} to N\mathbb{N}, which is well-ordered).
We claim that I=(d)I = (d). To prove this, we need to show that every element a in Ia \in I can be written as a=rda = rd for some r in Rr \in R.
Since d in Id \in I, clearly (d)sube I(d) \subseteq I.
Let a in Ia \in I. By the division algorithm in RR, there exist q,r in Rq, r \in R such that:
a=qd+r,a = qd + r,
where either r=0r = 0 or delta(r) < delta(d)\delta(r) < \delta(d).
Since a,d in Ia, d \in I and II is an ideal, qd in Iqd \in I. Thus, r=a-qd in Ir = a – qd \in I.
If r!=0r \neq 0, then delta(r) < delta(d)\delta(r) < \delta(d), which contradicts the minimality of delta(d)\delta(d). Hence, r=0r = 0, and a=qda = qd.
Therefore, a in(d)a \in (d), and we conclude I sube(d)I \subseteq (d).
Combining (d)sube I(d) \subseteq I and I sube(d)I \subseteq (d), we get I=(d)I = (d), showing that II is a principal ideal.
Conclusion:
Since every ideal in RR is principal, RR is a principal ideal domain. Thus, every Euclidean ring is a PID. “Q.E.D.”\boxed{\text{Q.E.D.}}
Question:-03
If W is any subspace of a finite dimensional inner product space V, prove that (W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W.
Answer:
Statement:
Let VV be a finite-dimensional inner product space, and let WW be a subspace of VV. Then:
(W^( _|_))^(_|_)=W.\left( W^\perp \right)^\perp = W.
Proof:
Definitions:
The orthogonal complementW^( _|_)W^\perp of a subspace W sube VW \subseteq V is defined as:
W^( _|_)={v in V:(:v,w:)=0″for all “w in W}.W^\perp = \{ v \in V : \langle v, w \rangle = 0 \ \text{for all } w \in W \}.
The space (W^( _|_))^(_|_)\left(W^\perp\right)^\perp is the orthogonal complement of W^( _|_)W^\perp, defined as:
(W^( _|_))^(_|_)={v in V:(:v,u:)=0″for all “u inW^( _|_)}.\left(W^\perp\right)^\perp = \{ v \in V : \langle v, u \rangle = 0 \ \text{for all } u \in W^\perp \}.
Key Observations:
Direct Sum Decomposition: In a finite-dimensional inner product space VV, any subspace W sube VW \subseteq V satisfies:
V=W o+W^( _|_),V = W \oplus W^\perp,
where:
W nnW^( _|_)={0}W \cap W^\perp = \{0\},
Every v in Vv \in V can be uniquely written as v=w+w^( _|_)v = w + w^\perp for w in Ww \in W and w^( _|_)inW^( _|_)w^\perp \in W^\perp.
Dimension Relation: The dimensions of WW, W^( _|_)W^\perp, and VV are related as:
Let v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perp. Then, vv satisfies (:v,u:)=0\langle v, u \rangle = 0 for all u inW^( _|_)u \in W^\perp.
By the direct sum decomposition V=W o+W^( _|_)V = W \oplus W^\perp, vv can be uniquely written as v=w+w^( _|_)v = w + w^\perp, where w in Ww \in W and w^( _|_)inW^( _|_)w^\perp \in W^\perp.
Since v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perp, we have: