Functional identities and their applications to graded algebras

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Examples of functional identities General FI Theory Applications

Functional identities and their applications to graded algebras

Matej Breˇsar,

University of Ljubljana, University of Maribor, Slovenia

July 2009

Matej Breˇsar Functional identities and their applications to graded algebras

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Example 1

R: a ring f :R→R

f(x)y =0 for all x,y ∈R

=⇒

f =0 orR “very special” (its left annihilator is nonzero: aR =0 witha6=0)

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Example 1

R: a ring

f :R→R

=⇒

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Example 1

R: a ring f :R→R

=⇒

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Example 1

R: a ring f :R→R

=⇒

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Example 1

R: a ring f :R→R

=⇒

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Example 1

R: a ring f :R→R

=⇒

f =0

orR “very special” (its left annihilator is nonzero: aR =0 witha6=0)

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Example 1

R: a ring f :R→R

=⇒

f =0 orR “very special”

(its left annihilator is nonzero: aR =0 witha6=0)

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Example 1

R: a ring f :R→R

=⇒

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Example 2

R prime (I,J ideals: IJ =0=⇒I =0 or J =0) f,g :R →R

f(x)y+g(y)x =0 for allx,y ∈R

=⇒

f =g =0 or (note: xy+ (−y)x =0 is an example)R is commutative

Proof.

f(x)(yz)w =−g(yz)xw =f(xw)yz =−g(y)xwz = f(x)ywz =⇒f(R)R[R,R] =0=⇒f =0 or[R,R] =0.

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Example 2

=⇒

Proof.

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Example 2

=⇒

Proof.

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Example 2

=⇒

Proof.

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Example 2

=⇒ f =g =0

or (note: xy+ (−y)x =0 is an example)R is commutative

Proof.

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Example 2

=⇒

f =g =0 or (note: xy+ (−y)x =0 is an example)

R is commutative

Proof.

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Example 2

=⇒

Proof.

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Example 2

=⇒

Proof.

f(x)(yz)w =−g(yz)xw

=f(xw)yz =−g(y)xwz = f(x)ywz =⇒f(R)R[R,R] =0=⇒f =0 or[R,R] =0.

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Example 2

=⇒

Proof.

f(x)(yz)w =−g(yz)xw =f(xw)yz =−g(y)xwz = f(x)ywz

=⇒f(R)R[R,R] =0=⇒f =0 or[R,R] =0.

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Example 2

=⇒

Proof.

f(x)(yz)w =−g(yz)xw =f(xw)yz =−g(y)xwz = f(x)ywz =⇒f(R)R[R,R] =0

=⇒f =0 or[R,R] =0.

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Example 2

=⇒

Proof.

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Example 3

Z: center ofR,R prime f,g :R →R

f(x)y+g(y)x ∈Z for all x,y ∈R

=⇒

f =g =0 or (note: x²−tr(x)x∈Z onR =M2(F),hence f(x)y+f(y)x ∈Z with f(x) =x−tr(x))R embeds in M₂(F)

Proof: Algebraic manipulations + structure theory of PI-rings

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Example 3

Z: center of R,R prime

f,g :R →R

=⇒

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Example 3

Z: center of R,R prime f,g :R →R

=⇒

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Example 3

=⇒

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Example 3

=⇒

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Example 3

=⇒ f =g =0

or (note: x²−tr(x)x∈Z onR =M2(F),hence f(x)y+f(y)x ∈Z with f(x) =x−tr(x))R embeds in M₂(F)

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Example 3

=⇒

f =g =0 or (note: x²−tr(x)x ∈Z onR =M2(F),

hence f(x)y+f(y)x ∈Z with f(x) =x−tr(x))R embeds in M₂(F)

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Example 3

=⇒

f =g =0 or (note: x²−tr(x)x ∈Z onR =M2(F),hence f(x)y+f(y)x ∈Z with f(x) =x−tr(x))

R embeds in M₂(F)

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Example 3

=⇒

f =g =0 or (note: x²−tr(x)x ∈Z onR =M2(F),hence f(x)y+f(y)x ∈Z with f(x) =x−tr(x))R embeds in M₂(F)

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Example 3

=⇒

Proof:

Algebraic manipulations + structure theory of PI-rings

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Example 3

=⇒

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Example 4

f₁,f₂,. . .,f_n :Rⁿ⁻¹ →R,R prime

f₁(x₂,. . .,x_n)x₁+f₂(x₁,x₃. . .,x_n)x₂+. . .+f_n(x₁,. . .,x_n₋₁)x_n ∈Z

=⇒

f1=f2 =. . .=fn=0 or R embeds inMn(F)

Note: A multilinear PI (polynomial identity) is such an FI withf_i

“polynomials”. FI theory: acomplement to PI theory.

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Example 4

f₁,f₂,. . .,f_n:Rⁿ⁻¹ →R,R prime

f₁(x₂,. . .,x_n)x₁+f₂(x₁,x₃. . .,x_n)x₂+. . .+f_n(x₁,. . .,x_n₋₁)x_n ∈Z

=⇒

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Example 4

f₁(x₂,. . .,x_n)x₁+f₂(x₁,x₃. . .,x_n)x₂+. . .+f_n(x₁,. . .,x_n₋₁)x_n∈ Z

=⇒

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Example 4

=⇒

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Example 4

=⇒

f1=f2 =_{. . .}=fn=0

or R embeds inMn(F) Note: A multilinear PI (polynomial identity) is such an FI withf_i

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Example 4

=⇒

f1=f2 =_{. . .}=fn=0 or R embeds inMn(F)

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Example 4

=⇒

“polynomials”.

FI theory: a complement to PI theory.

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Example 4

=⇒

“polynomials”. FI theory

: acomplement to PI theory.

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Example 4

=⇒

“polynomials”. FI theory: acomplementto PI theory.

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Example 5

f,g :R →R

f(x)y =xg(y) for all x,y ∈R

Expected solution:f(x) =xa,g(y) =ay for somea∈R. If 1∈ R: a=f(1) =g(1).

Without 1?E.g.,R is an ideal of S: thena may belong toS. In general: rings of quotients have to be involved.

In the context of prime rings,themaximal (left or right)ring of quotientsis suitable.

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Example 5

f,g :R →R

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Example 5

f,g :R →R

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Example 5

f,g :R →R

f(x)y =xg(y) for all x,y ∈R Expected solution:

f(x) =xa,g(y) =ay for somea∈R. If 1∈ R: a=f(1) =g(1).

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Example 5

f,g :R →R

Expected solution:f(x) =xa,g(y) =ay for somea∈R.

If 1∈ R: a=f(1) =g(1).

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1). Without 1?

E.g.,R is an ideal of S: thena may belong toS. In general: rings of quotients have to be involved.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

Without 1?E.g.,R is an ideal ofS:

thena may belong toS. In general: rings of quotients have to be involved.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

Without 1?E.g.,R is an ideal ofS: thena may belong toS.

In general: rings of quotients have to be involved.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

Without 1?E.g.,R is an ideal ofS: thena may belong toS. In general:

rings of quotients have to be involved.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

Without 1?E.g.,R is an ideal ofS: thena may belong toS. In general: rings of quotients have to be involved.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

In the context of prime rings,

themaximal (left or right)ring of quotientsis suitable.

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Example 5

f,g :R →R

If 1∈ R: a=f(1) =g(1).

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Example 6

f :R→R additive,R prime:

f(x)x =xf(x) for all x ∈R

=⇒

f(x) =λx+µ(x),

whereλ∈C, theextended centroid of R,andµ:R →C. (M. B., 1990)

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Example 6

f :R→R

additive,R prime:

=⇒

f(x) =λx+µ(x),

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Example 6

f :R→R additive,

R prime:

=⇒

f(x) =λx+µ(x),

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Example 6

=⇒

f(x) =λx+µ(x),

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Example 6

=⇒

f(x) =λx+µ(x),

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Example 6

=⇒

f(x) =λx+µ(x),

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Example 6

=⇒

f(x) =λx+µ(x),

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Example 6

=⇒

f(x) =λx+µ(x), whereλ∈C, theextended centroid of R,

andµ:R →C. (M. B., 1990)

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Example 6

=⇒

f(x) =λx+µ(x),

whereλ∈C, theextended centroid of R,and µ:R→C.

(M. B., 1990)

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Example 6

=⇒

f(x) =λx+µ(x),

whereλ∈C, theextended centroid of R,and µ:R→C. (M. B., 1990)

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Example 7

f :R×R→R biadditive,R prime:

f(x,x)x =xf(x,x) for all x∈R

=⇒

f(x,x) =λx²+µ(x)x+ν(x), whereλ∈C, and µ,ν:R →Cwith µadditive. (M. B., 1990)

Applications! Hint: interpretef(x,x)x =xf(x,x) as (x∗x)x =x(x∗x)

where∗is another (nonassociative) product on R.

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Example 7

f :R×R→R

biadditive,R prime:

=⇒

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Example 7

f :R×R→R biadditive,

R prime:

=⇒

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Example 7

=⇒

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Example 7

f(x,x)x =xf(x,x) for allx ∈R

=⇒

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Example 7

=⇒

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Example 7

=⇒

f(x,x) =λx²+µ(x)x+ν(x),

whereλ∈C, and µ,ν:R →Cwith µadditive. (M. B., 1990)

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Example 7

=⇒

f(x,x) =λx²+µ(x)x+ν(x), whereλ∈C, and µ,ν:R →C

with µadditive. (M. B., 1990)

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Example 7

=⇒

f(x,x) =λx²+µ(x)x+ν(x), whereλ∈C, and µ,ν:R →Cwith µadditive.

(M. B., 1990)

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Example 7

=⇒

(M. B., 1990)

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Example 7

=⇒

(M. B., 1990) Applications!

Hint: interpretef(x,x)x =xf(x,x) as (x∗x)x =x(x∗x)

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Example 7

=⇒

(M. B., 1990)

Applications! Hint: interpretef(x,x)x =xf(x,x)

as (x∗x)x =x(x∗x)

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Example 7

=⇒

(M. B., 1990)

Applications! Hint: interpretef(x,x)x =xf(x,x) as (x∗x)x= x(x∗x)

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Example 7

=⇒

(M. B., 1990)

Applications! Hint: interpretef(x,x)x =xf(x,x) as (x∗x)x= x(x∗x)

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Defining d -free sets

X: a subset of a ringQ with centerC

“Definition”(K. Beidar, M. Chebotar, 2000): X is a d-free subset of Q if FI’s such as

∑

d i=1

E_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d)x_i+x_iF_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d) =0 have onlystandard solutions, i.e.,

E_i =

∑

d

j=1 j6=i

x_jp_ij +λi, F_j =−

∑

d

i=1 i6=j

p_ijx_i−λj,

where

p_ij =p_ij(x₁,. . .,x_i₋₁,x_i+1,. . .,x_j₋_i,x_j+1,. . .,x_d)∈ Q, λ_i =λ_i(x1,. . .,x_i₋₁,x_i+1,. . .,x_d)∈C.

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Defining d -free sets

“Definition”(K. Beidar, M. Chebotar, 2000): X is a d-free subset of Q if FI’s such as

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

“Definition”(K. Beidar, M. Chebotar, 2000):

X is a d-free subset of Q if FI’s such as

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d)x_i+x_iF_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d) =0

have onlystandard solutions, i.e., E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d)x_i+x_iF_i(x₁,. . .,x_i₋₁,x_i₊₁,. . .,x_d) =0 have onlystandard solutions,

i.e., E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i =

∑

d

j=1 j6=i

x_jp_ij +λi,

F_j =−

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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Defining d -free sets

∑

d i=1

E_i =

∑

d

j=1 j6=i

∑

d

i=1 i6=j

p_ijx_i−λj,

where

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d -free sets exist!

K. Beidar (1998): A prime ringR is a d-free subset of its maximal left ring of quotientsQ

⇐⇒

R does not satisfy a PI of degree 2d −2 (i.e., R cannot be embedded inM_d₋₁(F)).

Other important examples ofd-free sets: Lie ideals of prime rings

symmetric elements of prime rings with involution

skew elements of prime rings with involution (and their Lie ideals)

semiprime rings

M_n(B),B any unital ring etc.