VMOU MT-01 SOLVED ASSIGNMENT | MA/M.SC. MT- 01(Advanced Algebra) | July-2024 & January-2025

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.
  1. (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 K H K H nn KH \cap KHK.
    (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.
  1. Prove that every Euclidean ring is a principal ideal domain.
  2. If W is any subspace of a finite dimensional inner product space V , the prove that ( W ) = W W = W (W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W(W)=W.
  3. Prove that every additive abelian group is a module over the ring Z of integers.
  4. If K K KKK is a field and d 1 , d 2 , . d n d 1 , d 2 , . d n d_(1),d_(2),dots.d_(n)d_1, d_2, \ldots . d_nd1,d2,.dn are distinct automorphisms of K K KKK. Then prove that it is impossible to find elements b 1 , b 2 , . b n b 1 , b 2 , . b n b_(1),b_(2),dots.b_(n)b_1, b_2, \ldots . b_nb1,b2,.bn not all zero in K K KKK such that b 1 ϕ 1 ( a ) + b 2 ϕ 2 ( a ) + + b n ϕ n ( a ) = 0 a K b 1 ϕ 1 ( a ) + b 2 ϕ 2 ( a ) + + b n ϕ n ( a ) = 0 a K 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 Kb1ϕ1(a)+b2ϕ2(a)++bnϕn(a)=0aK.
Section – C
(Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
  1. Let M 1 M 1 M_(1)M_1M1 and M 2 M 2 M_(2)M_2M2 are submodule of an R-module M . Then prove that:
M 1 + M 2 M 2 M 1 M 1 M 2 M 1 + M 2 M 2 M 1 M 1 M 2 (M_(1)+M_(2))/(M_(2))~=(M_(1))/(M_(1)nnM_(2))\frac{M_1+M_2}{M_2} \cong \frac{M_1}{M_1 \cap M_2}M1+M2M2M1M1M2
  1. Let V and V V V^(‘)V^{\prime}V be any two finite dimensional vector space over the same field F . Then prove that the vector space Hom ( V , V ) Hom V , V Hom(V,V^(‘))\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right)Hom(V,V) of all linear transformation of V to V V V^(‘)V^{\prime}V is also finite dimensional and dim Hom ( V , V ) = dim V × dim V Hom V , V = dim V × dim V Hom(V,V^(‘))=dimVxx dimV^(‘)\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right)=\operatorname{dim} \mathrm{V} \times \operatorname{dim} V^{\prime}Hom(V,V)=dimV×dimV.

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 K H K H nn KH \cap KHK.

Answer:

If G G GGG is the internal direct product of the subgroups H H HHH and K K KKK, the following conditions must hold:
  1. Every element of G G GGG can be uniquely written as a product h k h k hkh khk where h H h H h in Hh \in HhH and k K k K k in Kk \in KkK.
  2. H K = { e } H K = { e } H nn K={e}H \cap K = \{ e \}HK={e}, where e e eee is the identity element of G G GGG.
Thus, H K = { e } H K = { e } H nn K={e}H \cap K = \{ e \}HK={e}.

Question:-01(b)

Write class equation for the finite group.

Answer:

The class equation of a finite group G G GGG is a fundamental result in group theory that relates the structure of G G GGG to its conjugacy classes. It is expressed as:
| G | = | Z ( G ) | + i [ G : C G ( g i ) ] , | G | = | Z ( G ) | + i [ G : C G ( g i ) ] , |G|=|Z(G)|+sum_(i)[G:C_(G)(g_(i))],|G| = |Z(G)| + \sum_{i} [G : C_G(g_i)],|G|=|Z(G)|+i[G:CG(gi)],
where:
  • | G | | G | |G||G||G|: The order (size) of the group G G GGG,
  • Z ( G ) Z ( G ) Z(G)Z(G)Z(G): The center of G G GGG, i.e., the set of elements that commute with every element of G G GGG,
  • | Z ( G ) | | Z ( G ) | |Z(G)||Z(G)||Z(G)|: The size (order) of the center Z ( G ) Z ( G ) Z(G)Z(G)Z(G),
  • g i g i g_(i)g_igi: A representative of each conjugacy class outside the center,
  • C G ( g i ) C G ( g i ) C_(G)(g_(i))C_G(g_i)CG(gi): The centralizer of g i g i g_(i)g_igi in G G GGG, i.e., the set of elements in G G GGG that commute with g i g i g_(i)g_igi,
  • [ G : C G ( g i ) ] [ G : C G ( g i ) ] [G:C_(G)(g_(i))][G : C_G(g_i)][G:CG(gi)]: The index of the centralizer C G ( g i ) C G ( g i ) C_(G)(g_(i))C_G(g_i)CG(gi) in G G GGG, which equals the size of the conjugacy class containing g i g i g_(i)g_igi.

Interpretation:

  1. The elements in Z ( G ) Z ( G ) Z(G)Z(G)Z(G) form singleton conjugacy classes (since they commute with all elements).
  2. The remaining conjugacy classes contribute terms of the form [ G : C G ( g i ) ] [ G : C G ( g i ) ] [G:C_(G)(g_(i))][G : C_G(g_i)][G:CG(gi)], representing their sizes.
The class equation is a powerful tool to analyze the structure of G G GGG, 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 E E EEE be a field extension of F F FFF ( F E F E F sube EF \subseteq EFE). The extension E / F E / F E//FE/FE/F is called a normal extension if:
  1. Every irreducible polynomial f ( x ) F [ x ] f ( x ) F [ x ] f(x)in F[x]f(x) \in F[x]f(x)F[x] that has at least one root in E E EEE splits completely into linear factors over E E EEE, i.e., all the roots of f ( x ) f ( x ) f(x)f(x)f(x) are contained in E E EEE.

Equivalent Definitions:

  1. E / F E / F E//FE/FE/F is normal if E E EEE is the splitting field of some family of polynomials in F [ x ] F [ x ] F[x]F[x]F[x].
  2. E / F E / F E//FE/FE/F is normal if it is closed under conjugation by automorphisms, i.e., for any field automorphism σ σ sigma\sigmaσ of a larger field containing E E EEE, if σ ( e ) E σ ( e ) E sigma(e)in E\sigma(e) \in Eσ(e)E for some e E e E e in Ee \in EeE, then σ ( e ) σ ( e ) sigma(e)\sigma(e)σ(e) is also in E E EEE.

Examples:

  1. The extension Q ( 2 ) / Q Q ( 2 ) / Q Q(sqrt2)//Q\mathbb{Q}(\sqrt{2})/\mathbb{Q}Q(2)/Q is not normal, because the irreducible polynomial x 2 2 x 2 2 x^(2)-2x^2 – 2x22 over Q Q Q\mathbb{Q}Q has a root 2 2 sqrt2\sqrt{2}2 in Q ( 2 ) Q ( 2 ) Q(sqrt2)\mathbb{Q}(\sqrt{2})Q(2), but its other root, 2 2 -sqrt2-\sqrt{2}2, is not in Q ( 2 ) Q ( 2 ) Q(sqrt2)\mathbb{Q}(\sqrt{2})Q(2).
  2. The extension Q ( 2 , 3 3 ) / Q Q ( 2 , 3 3 ) / Q Q(sqrt2,root(3)(3))//Q\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})/\mathbb{Q}Q(2,33)/Q is normal, because it is the splitting field of ( x 2 2 ) ( x 3 3 ) ( x 2 2 ) ( x 3 3 ) (x^(2)-2)(x^(3)-3)(x^2 – 2)(x^3 – 3)(x22)(x33) over Q Q Q\mathbb{Q}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 u , v u,v\mathbf{u}, \mathbf{v}u,v in an inner product space, the following inequality holds:
| u , v | u v , | u , v | u v , |(:u,v:)| <= ||u||*||v||,|\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \cdot \|\mathbf{v}\|,|u,v|uv,
where:
  • u , v u , v (:u,v:)\langle \mathbf{u}, \mathbf{v} \rangleu,v: The inner product of u u u\mathbf{u}u and v v v\mathbf{v}v,
  • u = u , u u = u , u ||u||=sqrt((:u,u:))\|\mathbf{u}\| = \sqrt{\langle \mathbf{u}, \mathbf{u} \rangle}u=u,u: The norm (magnitude) of u u u\mathbf{u}u,
  • v = v , v v = v , v ||v||=sqrt((:v,v:))\|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}v=v,v: The norm (magnitude) of v v v\mathbf{v}v.

Equality Condition:

Equality holds if and only if u u u\mathbf{u}u and v v v\mathbf{v}v are linearly dependent, i.e., there exists a scalar λ λ lambda\lambdaλ such that u = λ v u = λ v u=lambdav\mathbf{u} = \lambda \mathbf{v}u=λv.

Special Cases:

  1. In Euclidean space R n R n R^(n)\mathbb{R}^nRn, with the standard dot product: | u v | u v , | u v | u v , |u*v| <= ||u||||v||,|\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\|,|uv|uv,where u = i = 1 n u i 2 u = i = 1 n u i 2 ||u||=sqrt(sum_(i=1)^(n)u_(i)^(2))\|\mathbf{u}\| = \sqrt{\sum_{i=1}^n u_i^2}u=i=1nui2.
  2. For functions f , g f , g f,gf, gf,g in an inner product space L 2 [ a , b ] L 2 [ a , b ] L^(2)[a,b]L^2[a, b]L2[a,b]: | a b f ( x ) g ( x ) d x | a b f ( x ) 2 d x a b g ( x ) 2 d x . a b f ( x ) g ( x ) d x a b f ( x ) 2 d x a b g ( x ) 2 d x . |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}.|abf(x)g(x)dx|abf(x)2dxabg(x)2dx.
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:

  1. Euclidean Ring: A commutative ring R R RRR with unity is a Euclidean ring if there exists a function δ : R { 0 } N δ : R { 0 } N delta:R\\{0}rarrN\delta: R \setminus \{0\} \to \mathbb{N}δ:R{0}N (called a Euclidean norm) such that for all a , b R a , b R a,b in Ra, b \in Ra,bR with b 0 b 0 b!=0b \neq 0b0:
    • There exist q , r R q , r R q,r in Rq, r \in Rq,rR such that a = b q + r a = b q + r a=bq+ra = bq + ra=bq+r, where either r = 0 r = 0 r=0r = 0r=0 or δ ( r ) < δ ( b ) δ ( r ) < δ ( b ) delta(r) < delta(b)\delta(r) < \delta(b)δ(r)<δ(b).
  2. Principal Ideal Domain (PID): A commutative ring R R RRR with unity is a PID if every ideal I R I R I sube RI \subseteq RIR is a principal ideal, i.e., there exists some a R a R a in Ra \in RaR such that I = ( a ) = { r a : r R } I = ( a ) = { r a : r R } I=(a)={ra:r in R}I = (a) = \{ ra : r \in R \}I=(a)={ra:rR}.

Proof:

Let R R RRR be a Euclidean ring, and let I I III be a nonzero ideal of R R RRR. We need to show that I I III is a principal ideal.
  1. Since I I III is nonzero, there exists some b I b I b in Ib \in IbI with b 0 b 0 b!=0b \neq 0b0. Among all nonzero elements of I I III, choose an element d I d I d in Id \in IdI such that δ ( d ) δ ( d ) delta(d)\delta(d)δ(d) is minimal (this is possible because δ δ delta\deltaδ maps elements of R { 0 } R { 0 } R\\{0}R \setminus \{0\}R{0} to N N N\mathbb{N}N, which is well-ordered).
  2. We claim that I = ( d ) I = ( d ) I=(d)I = (d)I=(d). To prove this, we need to show that every element a I a I a in Ia \in IaI can be written as a = r d a = r d a=rda = rda=rd for some r R r R r in Rr \in RrR.
    • Since d I d I d in Id \in IdI, clearly ( d ) I ( d ) I (d)sube I(d) \subseteq I(d)I.
    • Let a I a I a in Ia \in IaI. By the division algorithm in R R RRR, there exist q , r R q , r R q,r in Rq, r \in Rq,rR such that:
      a = q d + r , a = q d + r , a=qd+r,a = qd + r,a=qd+r,
      where either r = 0 r = 0 r=0r = 0r=0 or δ ( r ) < δ ( d ) δ ( r ) < δ ( d ) delta(r) < delta(d)\delta(r) < \delta(d)δ(r)<δ(d).
    • Since a , d I a , d I a,d in Ia, d \in Ia,dI and I I III is an ideal, q d I q d I qd in Iqd \in IqdI. Thus, r = a q d I r = a q d I r=a-qd in Ir = a – qd \in Ir=aqdI.
    • If r 0 r 0 r!=0r \neq 0r0, then δ ( r ) < δ ( d ) δ ( r ) < δ ( d ) delta(r) < delta(d)\delta(r) < \delta(d)δ(r)<δ(d), which contradicts the minimality of δ ( d ) δ ( d ) delta(d)\delta(d)δ(d). Hence, r = 0 r = 0 r=0r = 0r=0, and a = q d a = q d a=qda = qda=qd.
  3. Therefore, a ( d ) a ( d ) a in(d)a \in (d)a(d), and we conclude I ( d ) I ( d ) I sube(d)I \subseteq (d)I(d).
  4. Combining ( d ) I ( d ) I (d)sube I(d) \subseteq I(d)I and I ( d ) I ( d ) I sube(d)I \subseteq (d)I(d), we get I = ( d ) I = ( d ) I=(d)I = (d)I=(d), showing that I I III is a principal ideal.

Conclusion:

Since every ideal in R R RRR is principal, R R RRR is a principal ideal domain. Thus, every Euclidean ring is a PID. Q.E.D. Q.E.D. “Q.E.D.”\boxed{\text{Q.E.D.}}Q.E.D.

Question:-03

If W is any subspace of a finite dimensional inner product space V, prove that ( W ) = W W = W (W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W(W)=W.

Answer:

Statement:

Let V V VVV be a finite-dimensional inner product space, and let W W WWW be a subspace of V V VVV. Then:
( W ) = W . W = W . (W^( _|_))^(_|_)=W.\left( W^\perp \right)^\perp = W.(W)=W.

Proof:

Definitions:

  1. The orthogonal complement W W W^( _|_)W^\perpW of a subspace W V W V W sube VW \subseteq VWV is defined as:
    W = { v V : v , w = 0 for all w W } . W = { v V : v , w = 0 for all w W } . 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 \}.W={vV:v,w=0 for all wW}.
  2. The space ( W ) W (W^( _|_))^(_|_)\left(W^\perp\right)^\perp(W) is the orthogonal complement of W W W^( _|_)W^\perpW, defined as:
    ( W ) = { v V : v , u = 0 for all u W } . W = { v V : v , u = 0 for all u W } . (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 \}.(W)={vV:v,u=0 for all uW}.

Key Observations:

  1. Direct Sum Decomposition: In a finite-dimensional inner product space V V VVV, any subspace W V W V W sube VW \subseteq VWV satisfies:
    V = W W , V = W W , V=W o+W^( _|_),V = W \oplus W^\perp,V=WW,
    where:
    • W W = { 0 } W W = { 0 } W nnW^( _|_)={0}W \cap W^\perp = \{0\}WW={0},
    • Every v V v V v in Vv \in VvV can be uniquely written as v = w + w v = w + w v=w+w^( _|_)v = w + w^\perpv=w+w for w W w W w in Ww \in WwW and w W w W w^( _|_)inW^( _|_)w^\perp \in W^\perpwW.
  2. Dimension Relation: The dimensions of W W WWW, W W W^( _|_)W^\perpW, and V V VVV are related as:
    dim ( W ) + dim ( W ) = dim ( V ) . dim ( W ) + dim ( W ) = dim ( V ) . dim(W)+dim(W^( _|_))=dim(V).\dim(W) + \dim(W^\perp) = \dim(V).dim(W)+dim(W)=dim(V).
  3. Orthogonality Property: If v W v W v in Wv \in WvW, then v , u = 0 v , u = 0 (:v,u:)=0\langle v, u \rangle = 0v,u=0 for all u W u W u inW^( _|_)u \in W^\perpuW.

Proof Steps:

  1. Inclusion W ( W ) W ( W ) W sube(W^( _|_))^(_|_)W \subseteq (W^\perp)^\perpW(W):
    • Take any w W w W w in Ww \in WwW. For all u W u W u inW^( _|_)u \in W^\perpuW, by the definition of orthogonal complement: w , u = 0. w , u = 0. (:w,u:)=0.\langle w, u \rangle = 0.w,u=0.
    • Thus, w ( W ) w W w in(W^( _|_))^(_|_)w \in \left(W^\perp\right)^\perpw(W). Therefore, W ( W ) W W W sube(W^( _|_))^(_|_)W \subseteq \left(W^\perp\right)^\perpW(W).
  2. Inclusion ( W ) W ( W ) W (W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W(W)W:
    • Let v ( W ) v W v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perpv(W). Then, v v vvv satisfies v , u = 0 v , u = 0 (:v,u:)=0\langle v, u \rangle = 0v,u=0 for all u W u W u inW^( _|_)u \in W^\perpuW.
    • By the direct sum decomposition V = W W V = W W V=W o+W^( _|_)V = W \oplus W^\perpV=WW, v v vvv can be uniquely written as v = w + w v = w + w v=w+w^( _|_)v = w + w^\perpv=w+w, where w W w W w in Ww \in WwW and w W w W w^( _|_)inW^( _|_)w^\perp \in W^\perpwW.
    • Since v ( W ) v W v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perpv(W), we have:
      0 = v , w = w + w , w = w , w , 0 = v , w = w + w , w = w , w , 0=(:v,w^( _|_):)=(:w+w^( _|_),w^( _|_):)=(:w^( _|_),w^( _|_):),0 = \langle v, w^\perp \rangle = \langle w + w^\perp, w^\perp \rangle = \langle w^\perp, w^\perp \rangle,0=v,w=w+w,w=w,w,
      because w , w = 0 w , w = 0 (:w,w^( _|_):)=0\langle w, w^\perp \rangle = 0w,w=0 (orthogonality of W W WWW and W W W^( _|_)W^\perpW).
    • This implies w 2 = 0 w 2 = 0 ||w^( _|_)||^(2)=0\|w^\perp\|^2 = 0w2=0, so w = 0 w = 0 w^( _|_)=0w^\perp = 0w=0. Hence, v = w W v = w W v=w in Wv = w \in Wv=wW.
    • Therefore, ( W ) W ( W ) W (W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W(W)W.
  3. Conclusion:
    Combining W ( W ) W ( W ) W sube(W^( _|_))^(_|_)W \subseteq (W^\perp)^\perpW(W) and ( W ) W ( W ) W (W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W(W)W, we have:
    ( W ) = W . ( W ) = W . (W^( _|_))^(_|_)=W.(W^\perp)^\perp = W.(W)=W.
Q.E.D. Q.E.D. “Q.E.D.”\boxed{\text{Q.E.D.}}Q.E.D.

Question:-04

Prove that every additive abelian group is a module over the ring Z Z Z\mathbb{Z}Z of integers.

Answer:

Statement:

Every additive abelian group G G