Download Algebra Qualifying Exam Notes

Algebra Qualifying Exam Notes
Proposition 1: Let H be a subgroup of G of index p, where p is the smallest prime dividing the
order of G. Then H is a normal subgroup of G
Proposition 2: G/N is abelian ⇔ [G, G] ≤ N
Proof: (⇒) Let a, b ∈ G. Then [a, b]N = [aN, bN ] = N . So [a, b] ∈ N . Thus N contains all
commutators. So N contains [G, G], the subgroup generated by all commutators. Then, of course,
[G, G] ≤ N .
(⇐) Let a, b ∈ G. Then [aN, bN ] = [a, b]N = N , since [a, b] ∈ [G, G] ⊆ N . Therefore, G/N is
abelian.QED
Proposition 3: If H, K are subgroups of G and H ≤ NG (K), then HK is a subgroup of G
Corollary 4: If H, K are subgroups of G and one of them is normal in G, then HK is a subgroup
of G.
Second isomorphim theorem: If H ≤ G, N E G, then HN/N ∼
= H/(H ∩ N )
Proposition 5: If G is a simple group and H is a proper subgroup of G of index n, then there is
an injective homomorphisms φ : G → Sn .
Proof: Let G act on the cosets of H by left multiplication. This action permutes the cosets of H.
So it induces a homomorphism φ : G → S[G:H] ∼
= Sn . Note that φ is not the trivial homomorphism.
Otherwise, gH = H for all g ∈ G meaning that H = G. Therefore, ker φ 6= G. But ker φ is a
normal subgroup of G. Since G is simple, the only other option is for ker φ to be trivial. Therefore,
φ is injective. QED
Definition: The orbit of x is G · x = OrbG (x) := {g · x|g ∈ G}
Definition: The stabilizer of x is Gx = StabG (x) := {g ∈ G : g · x = x}
Orbit stabilizer theorem: |OrbG (x)| = [G : StabG (x)] = |G|/|StabG (x)|
Proof: Let φ : G/StabG (x) → OrbG (x) be defined by φ(gStabG (x)) = g · x. First let’s check
that this is well-defined. If gStabG (x) = g 0 StabG (x), then g = g 0 h for some h ∈ StabG (x). So
φ(gStabG (x)) = g · x = g 0 h · x = g 0 · x = φ(g 0 StabG (x)). Thus the map is well-defined. φ is clearly
surjective, since any element of OrbG (x) can be written in the form g · x = φ(gStabG (x)). Also
if φ(gStabG (x)) = φ(g 0 StabG (x)), then g · x = g 0 · x. So g −1 g 0 · x = x. Thus, g −1 g 0 ∈ StabG (x).
So gStabG (x) = gg −1 g 0 StabG (x) = g 0 StabG (x). Therefore, φ is also injective. So φ is a bijcetion.
Thus, |OrbG (x)| = |G/StabG (x)|. QED
1
Definition: The centralizer of a nonempty subset A of a group G is
CG (A) = {g ∈ G|gag −1 = a, ∀a ∈ A}. It is a subgroup of G.
Definition: The normalizer of a nonempty set A of a group G is
NG (A) = {g ∈ G|gAg −1 = A}. It is a subgroup of G.
Proposition 6: CG (A) ≤ NG (A)
Proposition 7: The number of conjugates of an element g ∈ G is [G : CG (g)]
Proof: This follows from the orbit-stabilizer theorem. QED
Sylow’s theorems
1. All Sylow p-subgroups are conjugate.
2. np (G)||G| and np (G) ≡ 1 mod p. In particular, np (G) = [G : NG (P )] where P is a Sylow
p-subgroup
Semidirect Products: Let G be a group. If N is a normal subgroup of G, H is a subgroup of G,
N H = G, and N ∩ H = {e}, then G ∼
= N o H.
Fratini’s Argument: Let G be a finite group. Let H be a normal subgroup of G, and let P be a
Sylow p-subgroup of H. Then G = HNG (P ) and |G : H| divides |NG (P )|.
Proof: Since H is normal in G, HNG (P ) is a subgroup of G. For any g ∈ G, gP g −1 ≤ gHg −1 = H.
Thus, gP g −1 is a Sylow p-subgroup of H. All Sylow p-subgroups of H are conjugate in H, so
there exists an h ∈ H such that hP h−1 = gP g −1 . Thus h−1 gP (h−1 g)−1 = P . So h−1 g ∈ NG (P ).
Therefore, g ∈ hNG (P ). So G ≤ HNG (P ). Therefore, G = HNG (P ).
By the second isomorphism theorem, [G : H] = |G/H| = |HNG (P )/H| = |NG (P )/(NG (P ) ∩ H| =
|NG (P )|/|NG (P ) ∩ H|. So [G : H] divides |NG (P )|. QED
Definition: The upper central series of a group G is the chain of subgroups 1 = Z0 (G) ≤ Z1 (G) ≤
Z2 (G) ≤ · · · such that Zi+1 (G)/Zi (G) = Z(G/Zi (G)).
Definition: The lower central series of a group G is the chaing of subgroups G0 ≥ G1 ≥ G2 ≥ · · · ,
where G0 = G, G1 = [G, G], and Gi+1 = [G, Gi ].
Definition: A group G is nilpotent if Zc (G) = G. (Equivalently, A group G is nilpotent if Gc = 1)
The smallest such c is the nilpotence class of G
Theorem 8: Let G be a finite group. The following are equivalent:
1. G is nilpotent
2. Every proper subgroup of G is a proper subgroup of its normalizer
2
3. Every Sylow Subgroup is normal in G
4. G is a direct product of its Sylow subgroups
5. Every maximal subgroup of G is normal.
Definition: The derived or commutator subgroup of a group G is [G, G]
Definition: The derived or commutator series of a group is the sequence
G(0) = G, G(1) = [G, G], G(i+1) = [G(i) , G(i) ]. The notation G0 = G(1) and G00 = G(2) is also used.
Definition: A group G is solvable if there is a chain of subgroups 1 = G0 / G1 / · · · / Gs = G Such
that Gi+1 /Gi is abelian for all i
Proposition 9: G is solvable ⇔ G(n) = 1 for some n ≥ 0.
Proposition 10: If N E G is solvable and G/N is solvable, then G is solvable.
Proposition 11: G(i) ≤ Gi for all i
Proposition 12: If G is nilpotent, then G is solvable.
Properties of Sn :
Proposition 13:
1. The transitive subgroups of S4 are S4 , A4 , D4 , V4 , C4 .
2. The transitive subgroups of S3 are S3 and A3 .
Proposition 14: For all n ≥ 3, n 6= 4, An is the only nontrivial proper normal subgroup of Sn .
For n = 4, Sn also has the normal subgroup {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} isomorphic to V4 .
Proof: (For n ≥ 5 only) Let H be a nontrivial proper normal subgroup of Sn . Then H ∩ An E An .
An is simple for n ≥ 5, so H ∩ An = An or 1. Suppose H ∩ An = 1. Since H, An are both
normal in Sn , we have that HAn is a supgroup of Sn and HAn /An = H/(H ∩ An ) ∼
= H. So
2 = |Sn /An | ≥ |HAn /An | = |H| > 1. Therefore, |H| = 2. So H = hσi for some nontrivial σ of order
2. Since H is also normal in Sn , conjugating σ by any element of Sn must be σ. Thus, σ ∈ Z(Sn ).
But Z(Sn ) = {e} for n ≥ 3. So we have a contradiction. Therefore, H ∩ An = An . So An ≤ H.
In fact, An E H, since An is normal in Sn . Thus, H/An ≤ Sn /An . So |H/An | ≤ |Sn /An | = 2.
Therefore, |H| = |An | or |H| = 2|An | = |Sn |. Thus, H = An or Sn . Since H is a proper subgroup
of Sn . H = An . QED
3
Proposition 15: Z(Sn ) = {e} for n ≥ 3
Proof: Let σ ∈ Sn \ {e}. Then there exists an i, j with i 6= j such that σ(i) = j. Let k ∈ {1, 2, ..., n}
with i 6= k 6= j. Then (i k)σ(i) = j 6= σ(k) = σ(i k)(i). Therefore, σ does not commute with (i k).
So σ ∈
/ Z(Sn ). Therefore, Z(Sn ) can at most contain the identity element, which it of course does.
Thus, Z(Sn ) = {e}. QED
Proposition 16: [Sn , Sn ] = An
Proof: This is trivial for n = 1, 2. For n ≥ 3, 6= 4, An is the only nontrivial proper normal
subgroups of Sn . [Sn , Sn ] is the smallest normal subgroup of Sn such that Sn /[Sn , Sn ] is abelian.
Sn /An ∼
= C2 is abelian. The only smaller subgroup of Sn is {e}, and Sn /{e} ∼
= Sn is not abelian.
Therefore, [Sn , Sn ] = An . Similarly, for n = 4, [Sn , Sn ] is either An or the normal subgroup
{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ∼
= V4 . [(1 2 3), (1 4)] = (1 2 3)(1 4)(1 3 2)(1 4) = (1 2 4).
Therefore, (1 2 4) ∈ [Sn , Sn ]. As (1 2 4) is not in the subgroup isomorphic to V4 , [Sn , Sn ] = An .
QED
Properties of An :
Proposition 17: An is a non-abelian simple group for all n ≥ 5
Proposition 18: A4 ∼
= V 4 o C3
Proposition 19: For all n ≥ 3, n 6= 4, An is a simple group. For n = 4, {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
isomorphic to V4 is the only nontirivial proper normal subgroup of An .
Proof: (For n = 4 only) I will show that the only nontrivial proper normal subgroup of V4 o C3 is
V4 . Let N 6= V4 be a nontrivial normal subgroup of V4 o C3 . N is not a subgroup of V4 (Simple
multiplications that N = V4 if that is the case). Therefore, N contains an element of the form vn
where v ∈ V4 , n ∈ C3 \ {0}. Such an element has order 3. The Sylow 3-subgroups of V4 o C3 have
order 3. Thus, N contains a Sylow 3-subgroup of V4 o C3 . Since Sylow 3-subgroups are conjugate
and N is normal, N conatins all Sylow 3-subgroups.n3 (V4 o C3 ) = 1 or 4. If n3 (V4 o C3 ) = 1, then
the semi-direct product would just be a direct product. Thus, n3 (V4 o C3 ) = 4. The intersection
of distinct Sylow 3-subgroups is trivial. Otherwise, they would be the same, since every nontrivial
element of a group of order 3 generates the group. Thus, there are 9 = 3 · 4 − 3 elements in total
in the Sylow 3-subgroups. So |N | ≥ 9 and |N | divides 12. Therefore, |N | = 12. So N = V4 o C3 .
Therefore, the only nontrivial subgroup of V4 o C3 not equal to V4 is the improper one. QED
Proposition 20: For n ≥ 4, Z(An ) = {e}
4
Proof: Z(An ) is a normal subgroup of An . For n ≥ 5, An is simple, so Z(An ) = An or {e}. But An is
not abelian for n ≥ 5, so Z(An ) = {e}. For n = 4, Z(An ) = {e} or {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},
since A4 is not abelian. (1 2)(3 4) and (1 2 3) don’t commute. Thus, Z(An ) = {e}. QED
Proposition 21: [An , An ] = An for n ≥ 5. [A4 , A4 ] = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ∼
= V4 .
Proof: [An , An ] is a normal subgroup of An . For n ≥ 5, An is simple, so [An , An ] = An of {e}. But
An is not abelian for n ≥ 5, so [An , An ] = An . For n = 4, [An , An ] = An , {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},
or {e}. [An , An ] 6= {e} because An is not abelian. An /{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ∼
= C3 ,
which is abelian. Since An is the only other normal subgroup of An , {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
is the smallest normal subgroup such that A/{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is abelian. Therefore, [A4 , A4 ] = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ∼
= V4 . QED
Definition: Aut(G) is the set of all automorphisms of a group G
Definition: Inn(G) is the set of all inner automorphisms. Those that arise as conjugation by an
element in G.
Definition: Out(G) = Aut(G)/Inn(G)
Proposition 22: Inn(G) ∼
= G/Z(G)
Definition: The characteristic polynomial of a matrix A is CA (x) = det(xI − A)
Definition: The minimal polynomial of a matrix A is the unique monic polynomial of smallest degree
which when evaluated at A is zero. It is denoted mA (x).
Proposition 23:
1. The characteristic polynomial of A is the product of the invariant factors of A
2. The minimal polynomial divides the characteristic polynomial of A
3. The minimal polynomial contains all the roots
4. The degree of the minimal polynomial of an n × n matrix A is at most n.
5. The degree of the characteristic polynomial of an n × n matrix A is n.
Definition: The invariant factors are the diagonal elements in the Smith Normal form with degree
at least one. The invariant factors a1 (x), ..., am (x) satisfy ai (x)|ai+1 (x). am (x) is the minimal
polynomial.
5




Pn

Definition: The companion matrix of an invariant factor a(x) = i=1 ai xi is 



0 ···
.
1 ..
.
0 ..
.. . .
.
.
0 ···
···
..
.
..
.
0
0
..
.
..
.
−a0

−a1
..
.








0 −an−2
1 −an−1
The rational canonical form of a matrix is the block diagonal matrix whose blocks are the companion
matrices of the invariant factors. If a matrix is in such a form, then the campanion matrices are
automomatically the companion matrices of the invariant factors.
Definition: The elementary divisors are obtained from the invariant factors by writing each invaraint
factors as the product of distinct linear factors to powers; these linear factors to their respective
powers are the elementary divisors. For example, if (x − 1)(x − 3)3 , (x − 1)(x − 2)(x − 3)3 , (x −
1)(x − 2)2 (x − 3)3 are the invariant factors, then the elementary divisors are (x − 1), (x − 3)3 , (x −
1), (x − 2), (x − 3)3 , (x − 1), (x − 2)2 , (x − 3)3





k
Definition: The Jordan block of (x − λ) is the k × k matrix 



λ
1
..
.
..
.
0
..
.
..
.
0 ···
0
..
.
..
.
..
.
···
...
..
.
..
.
..
.
0

0
.. 
. 


0 


1 
λ
The Jordan canonical form of a matrix is the block diagonal matrix whose blocks are the Jordan
blocks of the elementary divisors.
Proposition 24: Similar matrices have the same RCF and JCF
Proposition 25: A matrix is diagonalizable iff its minimal polynomial factors into linear factors
over the field.
Proof: If its minimal polynomial factors into linear factors over the field, then so do all of the
invariant factors. Thus all the elementary divisors are linear. Therefore, the Jordan canonical
form of the matrix is diagonal. So the matrix is diagonalizable. For the converse, if a matrix is
diagonalizable, its elementary divisors are all linear. So the invariant factors, in particular the
minimal polynomial, factor into linear factors over the field. QED
φ(n) = deg(Φn (x)) and cyclotomic polynomials
φ(pn1 1 pn2 2 · · · pnk k ) = (p1 )n1 −1 (p1 − 1)(p2 )n2 −1 (p2 − 1) · · · (pk )nk −1 (pk − 1), where the pi are distinct
primes.
Lemma 26: Over an algebraically closed field if W 6= 0 is an invariant subspace of V under B,
then W has an eigenvector in B.
6
Proof: There is a linear transormation T : W → W that corrsponds to left multiplication by B.
Choose a basis for W and extend it to a basis for V . Let B 0 be B with respect to this basis. The
matrix B 00 corresponding to T will be of smaller order than B 0 , because B 0 acts on a larger space.
In fact, B 00 is a submatrix of B 0 . The characteristic polynomial of B 00 has a root in the algebraically
closed field. Thus B 00 has an eigenvalue λ, and hence and eigenvector v ∈ W . Thus, T (v) = λv. So
Bv = T (v) = λv. Thus, B has an eigenvector in W .QED
Spectral Theorem:
1. Hermitian matrices (ones that equal their conjugate transpose) are diagonalizable over C
2. Symmetric matrices are diagonalizable over R
Proof: Let A be a hermitian matrix. Then A has an eigenvector v1 with eigenvalue λ. Let W1 be
the subspace of all vectors orthogonal to v1 . For any w ∈ V1 , hv1 , Awi = hAv1 , wi = λhv1 , wi = 0.
Thus, Aw ∈ W1 . Therefore, W1 is an invariant subspace of A. So A has an eigenvector v2 ∈ W1 .
Let W2 be the subspace of W1 consisting of all vectors orthogonal to v2 . Repreat this process, to get
a set
P of orthongal vectors. ThesePare necessarily linearly independent, so they form a basis for V .
(If ni=1 ai vi = 0, then aj = hvj , ni=1 ai vi i = hvj , 0i = 0.) With respect to this basis, A is diagonal.
The same holds for symmetric matrices. That is, they are diagonalizable over C. To show they are
diagonalizable over R, it is sufficient to show that the eigenvalues must all be real. Note that the
real component of any eigenvector is an eigenvector if this is the case. Let v be an eigenvector of
A with norm 1 and eigenvalue λ. Then λ = λhv, vi = hAv, vi = hv, Avi = hv, λvi = λ̄hv, vi = λ̄. So
λ is real. QED
Definition: M is a Noetherian R-module if any increasing chain of submmodules stabilizes.
Proposition 27: M is a Noetherian R-module iff every submodule of M is finitely generated.
Example of a finitely generated module that is not Noetherian: Q[x0 , x1 , ...] has submodule hx0 , x1 , ...i,
which is not finitely generated.
The dimension of a free module overP
a commutative
unity is invariant.
Pnring with P
m
Distribution with tensor products: ( i=1 Mi ) ⊗ ( j=1 Nj ) = i,j Mi ⊗ Nj
Definition: φ : M × N → L is R-balanced if it is bilinear and φ(mr, n) = φ(m, rn)
The universal property of tensor products: Let φ : M × N → L be an R-balanced map,
where R is a ring with unity, M is a right R-module, N is a left R-module, L is an R-moduule.
Then there is a unique R-module homomorphism Φ : M ⊗R N → L such that Φ(m ⊗ n) = φ(m, n).
Cm ⊗Z Cn ∼
= Cgcd(m,n) . Prove this using the map φ : Cm × Cn → Cgcd(m,n) defined by φ(a, b) = ab
mod gcd(m, n).
Chinese remainder theorem
7
1. Let n be a positive integer and let pn1 1 · · · pnk k be its factorization into powers of distinct primes.
n
Then Z/nZ ∼
= Z/pn1 1 Z × · · · × Z/pk k Z.
2. Let g(x) be a nonconstant monic polynomial in F [x], where F is a field, and let f1 (x)n1 · · · fk (x)nk
be its factorization into powers of distinct irrducible polynomials. Then F [x]/(g(x)) ∼
=
F [x]/(f1 (x)n1 ) × · · · × F [x]/(fk (x)nk ).
Definition: A module is simple if it has no nontrivial proper submodules.
Definition: A nonzero ring R with unity is semisimple if it is isomorphic to a finite direct sum of
simple modules.
Definition: Division rings are essentially fields with multiplication that is not commutative
Proposition 28: Finite division rings are fields
Artin-Wedderburn Theorem: Let R be a semisimple nonzero ring with unity (not necessarily
commutative). Then R ∼
= Πki=1 Mni (Di ) where the Di are division rings.
Proposition 29: PID ⇒ UFD. It follows that PID ⇒ Noetherian
Proposition 30:
1. R is an integral domain ⇒ R[x] is an integral domain
2. R is a UFD ⇔ R[x] is a UFD
3. R is a PID ⇒ R[x] is a UFD
4. F is a field ⇒ F [x] is a PID
Counterexample: Z is a PID, but Z[x] is not. Consider (2, x).
Definition: For any submodule N of M .
The annihilator of N is Ann(N ) = {r ∈ R : rn = 0, ∀n ∈ N }. It is an ideal of R.
P
Eisenstein Criterion: Let p ∈ Z be prime. f = ni=1 ai xi ∈ Z[x] is Eisenstein if an ∈
/ pZ, all
other ai ∈ pZ and a0 ∈
/ p2 Z. If f is Eisenstein, it is irreducible in Q[x]. If f is also primitive (i.e.
the gcd of its coefficients is 1), then it is irreducible in Z[x].
Proposition 31: F (α)/F is algebraic iff α is algebraic over F .
Definition:
1. A polynomial is separable if all of its roots are have multiplicity 1.
2. α ∈ E is separable over F if its minimal polynomial is separable.
3. E/F is separable if every α ∈ E is separable over F .
8
Proposition 32: Let K/F be algebraic. Then K/F is separable iff K/E and E/F are separable.
Primitive Element Theorem: EVery finite separable field extension E/F is simple. That is,
E = F (α).
Definition: E/F is normal if it is the splitting field of some set of polynomials.
Proposition 33: Let K/E/F . Then K/F normal implies that K/E is normal
Proposition 34: E/F and K/F normal implies EK/F normal
Definition: A field extension K/F is Galois if it is both normal and separable
Proposition 35: K/F is Galois iff K is the splitting field of some separable polynomial over F .
Proposition 36: If f (x) ∈ Q[x] has degree n and K is the splitting field of f over Q, then
Gal(K/Q) is isomorphic to a transitive subrgroup of Sn .
Proposition 37: |Gal(E/F )| = [E : F ]
Definition: For G ≤ Gal(E/F ), E G is the largest subfield of E fixed by G.
Proposition 38: Gal(K/K H ) = H
Fundamental Theorem of Galois Theory: There is a one to one correspondence of intermediate
fields of K/F and subgroups of Gal(K/F ) with normal extensions E/F corresponding to normal
subgroups. H ≤ Gal(K/F ) corresponds to K H .
Proposition 39: Gal(Q(ζn )/Q) ∼
= (Z/nZ)× , where ζn is a primitive nth root of unity.
Proposition 40:
1. If p is an odd prime, then (Z/pn Z)× ∼
= Z/(p − 1)Z × Z/pn−1 Z.
2. If p is an even prime, then (Z/pn Z)× ∼
= Z/pZ × Z/pn−2 Z
Proposition 41:
n
1. Fpn is a field of roots of xp − x.
2. All finite fields are of the form Fpn .
9
3. The Galois group of Fpn /Fp is cyclic of order n and is generated by the Frobenius map:
x 7→ xp .
4. Fpn ⊆ Fpm ⇔ n|m
Proposition 42: The squareroots of distinct primes are linearly independent.
Descartes rule of signs: Roots with multiplicity great than 1, are counted multiple times.
1. Positive roots = # of sign changes of coefficients or some multiple of 2 less
2. Negative roots = # of sign changes of coefficients or some multiple of 2 less after multiplying
odd terms by -1
Cramer’s rule: If Av = B where A is an n × n matrix and v, B are vectors, then vi det A = det A0
where A0 is the matrix A with the ith column replaced by B.
1
1
···
1
1 x1
x2 · · · xn−1 xn x2
x22 · · · x2n−1 x2n = Π1≤i<j≤n (xj − xi )
1
Vandermonde determinant formula: ..
..
..
..
.. .
.
.
.
. n−1
n−1
n−1
n−1
x
x
··· x
x
Proof:
Subtract the
1
1
···
x1
x2 · · ·
x2
x22 · · ·
1
..
..
..
.
.
.
n−1
n−1
x
·
·
·
x
2
1
n−1
2
1
nth column
1
1
xn−1 xn
x2n−1 x2n
..
..
.
.
xn−1
n−1
from
all the other columns. To get
0
0
···
x1 − xn
x2 − xn
···
2 − x2
x2 − x2
x
···
n
n
1
2
=
.
.
..
.
..
.
n−1 .
xn−1 x
− xn−1 xn−1 − xn−1 · · ·
n
n
1
Multiplty the (n − 1)th row by −xn and add it to the n
(i − 1)th row by −xn and add it to the i row. This gets us
0
0
···
0
x
−
x
x
−
x
·
·
·
x
− xn
1
n
2
n
n−1
(x1 − xn )x1
(x
−
x
)x
·
·
·
(x
−
x
2
n 2
n−1
n )xn−1
=
.
.
.
.
..
..
..
..
(x1 − xn )xn−2 (x2 − xn )xn−2 · · · (xn−1 − xn )xn−1
1
2
n−1
Pulling out the factors of xi − xn we get
0
0
1
1
x1
n−1
x
2
= Π1≤i<n (xn −xi )×(−1)
..
..
.
.
n−2
n−2
x
x
1
2
n
2
···
···
···
..
.
xn−1
..
.
···
n−1
xn−1
0
1
n
0
xn−1 − xn
x2n−1 − x2n
..
.
1
xn
x2n
..
.
n−1 xn−1
xn−1
n
n−1 − xn
row. Repeat this by multipltying the
0 1
0
0
..
.
1
1
···
x1
x
···
2
= Π1≤i<n (xn −xi )× ..
..
..
.
.
.
n−2
n−2
x
x2
···
1
0 1
0
0
..
.
So the Vandermonde determinanat formula is true for an n × n matrix if it is true for an (n − 1) ×
(n − 1) matrix. It is also true for n = 2. So by induction it is true for all n ≥ 2. QED
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1
xn−1
..
.
xn−1
n−1
Definition: An R-algebra A is a ring with unity along with multiplication by elements in R and this
multiplication is commutative: r · a = a · r.
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