aarjaneiro/planetmath huggingface dataset:
support (graded ring)

support (graded ring)

Let

S

be a groupoid (semigroup, group) and suppose

R

=

⊕

s

∈

S

R

s

is an

S

-

graded ring

.

We define the

support

of

r

to be the set

supp

⁢

(

r

)

=

{

s

∈

S

∣

r

s

≠

0

}

. We can extend this definition to

supp

(

R

)

=

⋃

supp

R

s

≠

0

}

. If

supp

⁢

(

R

)

is a

finite set

then we say that the ring

R

has

finite
support

.

Title

support (graded ring)

Canonical name

SupportgradedRing

Date of creation

2013-03-22 15:40:00

Last modified on

2013-03-22 15:40:00

Owner

aplant (12431)

Last modified by

aplant (12431)

Numerical id

6

Author

aplant (12431)

Entry type

Definition

Classification

msc 13A02

Related topic

GradedRing

Defines

support

Defines

finite support

Generated on Fri Feb 9 17:19:42 2018 by

LaTeXML
any divisor is gcd of two principal divisors

any divisor is gcd of two principal divisors

Using the

exponent valuations

, one can easily prove the

Theorem

.

In any

divisor theory

, each

divisor

is the

greatest common divisor

of two principal divisors.

Proof.

Let

𝒪

*

→

𝔇

be a divisor theory and

𝔡

an arbitrary divisor in

𝔇

.  We may suppose that

𝔡

is not a principal divisor (if

𝔇

contains exclusively principal divisors, then

𝔡

=

gcd

⁡

(

𝔡

,

𝔡

)

and the proof is ready).  Let

𝔡

=

∏

i

=

1

r

𝔭

i

k

i

where the

𝔭

i

’s are pairwise distinct

prime divisors

and every

k

i

>

0

.  Then third condition in the theorem concerning

divisors and exponents

allows to choose an

element

α

of the ring

𝒪

such that

ν

𝔭

1

⁢

(

α

)

=

k

1

,

…

,

ν

𝔭

r

⁢

(

α

)

=

k

r

.

Let the principal divisor corresponding to

α

be

(

α

)

=

∏

i

=

1

r

𝔭

i

k

i

⁢

∏

j

=

1

s

𝔮

j

l

j

=

𝔡

⁢

𝔡

′

,

where the prime divisors

𝔮

j

are pairwise different among themselves and with the divisors

𝔭

i

.  We can then choose another element

β

of

𝒪

such that

ν

𝔭

1

⁢

(

β

)

=

k

1

,

…

,

ν

𝔭

r

⁢

(

β

)

=

k

r

,

ν

𝔮

1

⁢

(

β

)

=

…

=

ν

𝔮

s

⁢

(

β

)

=

0

.

Then we have

(

β

)

=

𝔡

⁢

𝔡

′′

,  where

𝔡

′′

∈

𝔇

and

gcd

⁡

(

𝔡

′

,

𝔡

′′

)

=

𝔮

0

⁢

⋯

⁢

𝔮

0

=

𝔢

=

(

1

)

.

The gcd of the principal divisors

(

α

)

and

(

β

)

is apparently

𝔡

, whence the proof is settled.

Title

any divisor is gcd of two principal divisors

Canonical name

AnyDivisorIsGcdOfTwoPrincipalDivisors

Date of creation

2013-03-22 17:59:37

Last modified on

2013-03-22 17:59:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A05

Classification

msc 13A18

Classification

msc 12J20

Related topic

TwoGeneratorProperty

Related topic

SumOfIdeals

Generated on Fri Feb 9 17:19:46 2018 by

LaTeXML
divisibility in rings

divisibility in rings

Let

(

A

,

+

,

⋅

)

be a

commutative ring

with a

non-zero
unity

1.  If

a

and

b

are two elements of

A

and if there
is an

element

q

of

A

such that

b

=

q

⁢

a

,  then

b

is
said to be

divisible

by

a

; it may be denoted by

a

∣

b

.  (If

A

has no

zero divisors

and

a

≠

0

,
then

q

is uniquely determined.)

When

b

is divisible by

a

,

a

is said to be a

divisor

or

factor (

http://planetmath.org/DivisibilityInRings

)

of

b

.  On the other hand,

b

is not said to be
a

multiple

of

a

except in the case that

A

is the
ring

ℤ

of the integers.  In some

languages

, e.g. in
the Finnish,

b

has a name which could be approximately be
translated as ‘

containant

’:

b

is a

containant

of

a

(“

b

on

a

:n

sisältäjä

”).

•

a

∣

b

iff

(

b

)

⊆

(

a

)

[see the

principal ideals

].

•

Divisibility is a

reflexive

and

transitive relation

in

A

.

•

0 is divisible by all elements of

A

.

•

a

∣

1

iff

a

is a unit of

A

.

•

All elements of

A

are divisible by every unit of

A

.

•

If

a

∣

b

then

a

n

∣

b

n

(

n

=

1

,

2

,

…

)

.

•

If

a

∣

b

then

a

∣

b

c

and

a

c

∣

b

c

.

•

If

a

∣

b

and

a

∣

c

then

a

∣

b

+

c

.

•

If

a

∣

b

and

a

∤

c

then

a

∤

b

+

c

.

Note.

The divisibility can be similarly defined if

(

A

,

+

,

⋅

)

is only a

semiring

; then it also has the
above properties except the first.  This concerns especially
the case that we have a ring

R

with non-zero unity and

A

is
the set of the ideals of

R

(see the ideal

multiplication

laws).
Thus one may speak of the

divisibility of ideals

in

R

:

𝔞

∣

𝔟

⇔

(

∃

𝔮

)

(

𝔟

=

𝔮

𝔞

)

.  Cf.

multiplication ring

.

Title

divisibility in rings

Canonical name

DivisibilityInRings

Date of creation

2015-05-06 15:18:14

Last modified on

2015-05-06 15:18:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

24

Author

pahio (2872)

Entry type

Definition

Classification

msc 13A05

Classification

msc 11A51

Related topic

PrimeElement

Related topic

Irreducible

Related topic

GroupOfUnits

Related topic

DivisibilityByPrimeNumber

Related topic

GcdDomain

Related topic

CorollaryOfBezoutsLemma

Related topic

ExistenceAndUniquenessOfTheGcdOfTwoIntegers

Related topic

MultiplicationRing

Related topic

IdealDecompositionInDedekindDomain

Related topic

IdealMultiplicationLaws

Related topic

UnityPlusNilpotentIsUnit

Defines

divisible

Defines

divisibility

Defines

divisibility of ideals

Generated on Fri Feb 9 17:19:50 2018 by

LaTeXML
divisor as factor of principal divisor

divisor as factor of principal divisor

Let an

integral domain

𝒪

have a

divisor theory

𝒪

*

→

𝔇

.  The definition of divisor theory (

http://planetmath.org/DivisorTheory

) implies that for any

divisor

𝔞

, there exists an element

ω

of

𝒪

such that

𝔞

divides the principal divisor

(

ω

)

, i.e. that

𝔞

⁢

𝔠

=

(

ω

)

with

𝔠

a divisor.  The following theorem states that

𝔠

may always be chosen such that it is

coprime

with any beforehand given divisor.

Theorem.

For any two divisors

𝔞

and

𝔟

, there is a principal divisor

(

ω

)

such that

𝔞

⁢

𝔠

=

(

ω

)

and

gcd

⁡

(

𝔟

,

𝔠

)

=

(

1

)

.

Proof.

Let

𝔭

1

,

…

,

𝔭

s

all distinct

prime divisors

, which divide the product

𝔞

⁢

𝔟

, and let the divisor

𝔞

be exactly

divisible

(

http://planetmath.org/ExactlyDivides

) by the powers

𝔭

1

a

1

,

…

,

𝔭

s

a

s

(the cases

a

i

=

0

are not excluded).  For each

i

=

1

,

…

,

s

,  we choose a nonzero element

α

i

of

𝒪

being exactly divisible by the power

𝔭

i

a

i

; the choosing is possible, since any nonzero element of the ideal determined by the divisor

𝔭

i

a

i

, not belonging to the sub-ideal determined by the divisor

𝔭

i

a

i

+

1

, will do.  According to the

Chinese remainder theorem

(

http://planetmath.org/ChineseRemainderTheoremInTermsOfDivisorTheory

), there exists a nonzero element

ω

of the ring

𝒪

such that

ω

≡

α

i

mod

𝔭

i

a

i

+

1

(

i

=

1

,

…

,

s

)

.

(1)

Because

α

i

is divisible by

𝔭

i

a

i

, the element

ω

is divisible by

𝔭

1

a

1

⁢

⋯

⁢

𝔭

s

a

s

=

𝔞

,  i.e.

(

ω

)

=

𝔞

⁢

𝔠

.  If one of the divisors

𝔭

i

would divide

𝔠

, then

(

ω

)

would be divisible by

𝔭

i

a

i

+

1

and thus by (1), also

α

i

were divisible by

𝔭

i

a

i

+

1

.  Therefore, no one of the prime divisors

𝔭

1

,

…

,

𝔭

s

divides

𝔠

.  On the other hand, every prime divisor dividing the divisor

𝔟

divides

𝔞

⁢

𝔟

and thus is one of

𝔭

1

,

…

,

𝔭

s

.  Accordingly, the divisors

𝔟

and

𝔠

have no common prime divisor, i.e.

gcd

⁡

(

𝔟

,

𝔠

)

=

(

1

)

.

References

1

М. М. Постников:

Введение  в  теорию  алгебраических
чисел

.  Издательство
‘‘Наука’’. Москва (1982).

Title

divisor as factor of principal divisor

Canonical name

DivisorAsFactorOfPrincipalDivisor

Date of creation

2013-03-22 18:02:09

Last modified on

2013-03-22 18:02:09

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A05

Classification

msc 11A51

Related topic

EveryIdealInADedekindDomainIsAFactorOfAPrincipalIdeal

Generated on Fri Feb 9 17:20:03 2018 by

LaTeXML
Eisenstein criterion

Eisenstein criterion

Theorem

(Eisenstein criterion).

Let

f

be a

primitive polynomial

over a

commutative

unique factorization domain

R

, say

f

⁢

(

x

)

=

a

0

+

a

1

⁢

x

+

a

2

⁢

x

2

+

…

+

a

n

⁢

x

n

.

If

R

has an

irreducible element

p

such that

p

∣

a

m

0

≤

m

≤

n

-

1

p

2

∤

a

0

p

∤

a

n

then

f

is

irreducible

.

Proof.

Suppose

f

=

(

b

0

+

…

+

b

s

⁢

x

s

)

⁢

(

c

0

+

…

+

c

t

⁢

x

t

)

where

s

>

0

and

t

>

0

. Since

a

0

=

b

0

⁢

c

0

, we know that

p

divides one but not both of

b

0

and

c

0

; suppose

p

∣

c

0

. By

hypothesis

, not all the

c

m

are divisible by

p

; let

k

be the smallest index such that

p

∤

c

k

. We have

a

k

=

b

0

⁢

c

k

+

b

1

⁢

c

k

-

1

+

…

+

b

k

⁢

c

0

.
We also have

p

∣

a

k

, and

p

divides every summand except one on the right side, which yields a

contradiction

. QED
∎

Title

Eisenstein criterion

Canonical name

EisensteinCriterion

Date of creation

2013-03-22 12:16:32

Last modified on

2013-03-22 12:16:32

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

13

Author

Daume (40)

Entry type

Theorem

Classification

msc 13A05

Synonym

Eisenstein irreducibility criterion

Related topic

GausssLemmaII

Related topic

IrreduciblePolynomial2

Related topic

Monic2

Related topic

AlternativeProofThatSqrt2IsIrrational

Generated on Fri Feb 9 17:20:06 2018 by

LaTeXML
Eisenstein criterion in terms of divisor theory

Eisenstein criterion in terms of divisor theory

The below

theorem

generalises

Eisenstein criterion

of irreducibility from UFD’s to domains with

divisor theory

.

Theorem

.

Let

f

⁢

(

x

)

:=

a

0

+

a

1

⁢

x

+

…

+

a

n

⁢

x

n

be a

primitive polynomial

over an

integral domain

𝒪

with divisor theory (

http://planetmath.org/DivisorTheory

)

𝒪

*

→

𝔇

.  If there is a

prime divisor

𝔭

∈

𝔇

such that

•

𝔭

∣

a

0

,

a

1

,

…

,

a

n

-

1

,

•

𝔭

∤

a

n

,

•

𝔭

2

∤

a

0

,

then the

polynomial

is

irreducible

.

Proof.

Suppose that we have in

𝒪

⁢

[

x

]

the factorisation

f

⁢

(

x

)

=

(

b

0

+

b

1

⁢

x

+

…

+

b

s

⁢

x

s

)

⁢

(

c

0

+

c

1

⁢

x

+

…

+

c

t

⁢

x

t

)

with

s

>

0

and

t

>

0

.  Because the principal divisor

(

a

0

)

, i.e.

(

b

0

)

⁢

(

c

0

)

is divisible by the prime divisor

𝔭

and there is a unique factorisation in the monoid

𝔇

,

𝔭

must divide

(

b

0

)

or

(

c

0

)

but, by

𝔭

2

∤

(

a

0

)

, not both of

(

b

0

)

and

(

c

0

)

; suppose e.g. that

𝔭

∣

c

0

.  If

𝔭

would divide all the

coefficients

c

j

, then it would divide also the

product

b

s

⁢

c

t

=

a

n

.  So, there is a certain smallest index

k

such that

p

∤

c

k

.  Accordingly, in the sum

b

0

⁢

c

k

+

b

1

⁢

c

k

-

1

+

…

+

b

k

⁢

c

0

, the prime divisor

𝔭

divides (

http://planetmath.org/DivisibilityInRings

) every summand except the first (see the definition of divisor theory (

http://planetmath.org/DivisorTheory

)); therefore it cannot divide the sum.  But the value of the sum is

a

k

which by

hypothesis

is divisible by the prime divisor.  This

contradiction

shows that the polynomial

f

⁢

(

x

)

is irreducible.

Title

Eisenstein criterion in terms of divisor theory

Canonical name

EisensteinCriterionInTermsOfDivisorTheory

Date of creation

2013-03-22 18:00:45

Last modified on

2013-03-22 18:00:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A05

Related topic

DivisorTheory

Generated on Fri Feb 9 17:20:12 2018 by

LaTeXML
example of using Eisenstein criterion

example of using Eisenstein criterion

For showing the irreducibility (

http://planetmath.org/IrreduciblePolynomial2

) of the

polynomial

P

⁢

(

x

)

:=

x

5

+

5

⁢

x

+

11

one would need a

prime number

dividing its other coefficients except the first one, but there is no such prime.  However, a suitable

x

:=

y

+

a

may change the situation.  Since  the

binomial coefficients

of

(

y

-

1

)

5

except the first and the last one are

divisible

by 5 and

11

≡

1

(

mod

5

)

,  we try

x

:=

y

-

1

.

Then

P

⁢

(

y

-

1

)

=

y

5

-

5

⁢

y

4

+

10

⁢

y

3

-

10

⁢

y

2

+

10

⁢

y

+

5

.

Thus the prime 5 divides other coefficients except the first one and the square of 5 does not divide the constant term of this polynomial in

y

, whence the

Eisenstein criterion

says that

P

⁢

(

y

-

1

)

is

irreducible

(in the field

ℚ

of its coefficients).  Apparently, also

P

⁢

(

x

)

must be irreducible.

It would be easy also to see that

P

⁢

(

x

)

does not have

rational

zeroes (

http://planetmath.org/RationalRootTheorem

).

Title

example of using Eisenstein criterion

Canonical name

ExampleOfUsingEisensteinCriterion

Date of creation

2013-03-22 19:10:14

Last modified on

2013-03-22 19:10:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Example

Classification

msc 13A05

Classification

msc 11C08

Generated on Fri Feb 9 17:20:15 2018 by

LaTeXML
primal element

primal element

An element

r

in a

commutative ring

R

is called

primal

if whenever

r

∣

a

b

, with

a

,

b

∈

R

, then there
exist elements

s

,

t

∈

R

such that

1.

r

=

s

⁢

t

,

2.

s

∣

a

and

t

∣

b

.

Lemma

. In a commutative ring, an element that is both

irreducible

and primal is a

prime element

.

Proof.

Suppose

a

is irreducible and primal, and

a

∣

b

c

. Since

a

is primal, there is

x

,

y

∈

R

such that

a

=

x

⁢

y

, with

x

∣

b

and

y

∣

c

. Since

a

is irreducible, either

x

or

y

is a unit. If

x

is a unit, with

z

as its

inverse

, then

z

⁢

a

=

z

⁢

x

⁢

y

=

y

, so that

a

∣

y

. But

y

∣

c

, we have that

a

∣

c

.
∎

Title

primal element

Canonical

name

PrimalElement

Date of creation

2013-03-22 14:50:21

Last modified on

2013-03-22 14:50:21

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13A05

Defines

primal

Generated on Fri Feb 9 17:20:19 2018 by

LaTeXML
Hilbert’s Nullstellensatz

Hilbert’s Nullstellensatz

Let

K

be an algebraically closed field, and let

I

be an ideal in

K

⁢

[

x

1

,

…

,

x

n

]

, the

polynomial ring

in

n

indeterminates.

Define

V

⁢

(

I

)

, the

zero set

of

I

, by

V

⁢

(

I

)

=

{

(

a

1

,

…

,

a

n

)

∈

K

n

∣

f

⁢

(

a

1

,

…

,

a

n

)

=

0

⁢

for all

⁢

f

∈

I

}

Weak Nullstellensatz

:

If

V

⁢

(

I

)

=

∅

, then

I

=

K

⁢

[

x

1

,

…

,

x

n

]

. In other words, the zero set of any

proper ideal

of

K

⁢

[

x

1

,

…

,

x

n

]

is nonempty.

Hilbert’s (Strong) Nullstellensatz

:

Suppose

f

∈

K

⁢

[

x

1

,

…

,

x

n

]

satisfies

f

⁢

(

a

1

,

…

,

a

n

)

=

0

for every

(

a

1

,

…

,

a

n

)

∈

V

⁢

(

I

)

. Then

f

r

∈

I

for some

integer

r

>

0

.

In the of

algebraic geometry

, the latter result is equivalent to the statement that

Rad

⁡

(

I

)

=

I

⁢

(

V

⁢

(

I

)

)

, that is, the

radical

of

I

is equal to the ideal of

V

⁢

(

I

)

.

Title

Hilbert’s Nullstellensatz

Canonical name

HilbertsNullstellensatz

Date of creation

2013-03-22 13:03:59

Last modified on

2013-03-22 13:03:59

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

8

Author

rmilson (146)

Entry type

Theorem

Classification

msc 13A10

Synonym

Nullstellensatz

Related topic

RadicalOfAnIdeal

Related topic

AlgebraicSetsAndPolynomialIdeals

Defines

zero set

Defines

Hilbert’s Nullstellensatz

Defines

weak Nullstellensatz

Generated on Fri Feb 9 17:20:24 2018 by

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nilradical

nilradical

Let

R

be a

commutative ring

. An

element

x

∈

R

is said to be

nilpotent

if

x

n

=

0

for some

positive integer

n

. The set of all

nilpotent elements

of

R

is an ideal of

R

, called the

nilradical

of

R

and denoted

Nil

⁡

(

R

)

. The nilradical is so named because it is the

radical

of the

zero ideal

.

The nilradical of

R

equals the

prime radical

of

R

, although proving that the two are

equivalent

requires the

axiom of choice

.

Title

nilradical

Canonical name

Nilradical

Date of creation

2013-03-22 12:47:52

Last modified on

2013-03-22 12:47:52

Owner

djao (24)

Last modified by

djao (24)

Numerical id

4

Author

djao (24)

Entry type

Definition

Classification

msc 13A10

Related topic

PrimeRadical

Related topic

JacobsonRadical

Defines

nilpotent

Generated on Fri Feb 9 17:20:27 2018 by

LaTeXML
proof of Hilbert’s Nullstellensatz

proof of Hilbert’s Nullstellensatz

Let

K

be an algebraically closed field, let

n

≥

0

, and let

I

be an ideal of the

polynomial ring

K

⁢

[

x

1

,

…

,

x

n

]

. Let

f

∈

K

⁢

[

x

1

,

…

,

x

n

]

be
a

polynomial

with the property that

f

⁢

(

a

1

,

…

,

a

n

)

=

0

⁢

for all

⁢

(

a

1

,

…

,

a

n

)

∈

V

⁢

(

I

)

.

Suppose that

f

r

∉

I

for all

r

>

0

; in particular,

I

is strictly smaller than

K

⁢

[

x

1

,

…

,

x

n

]

and

f

≠

0

. Consider the ring

R

=

K

⁢

[

x

1

,

…

,

x

n

,

1

/

f

]

⊂

K

⁢

(

x

1

,

…

,

x

n

)

.

The

R

-ideal

R

⁢

I

is strictly smaller than

R

, since

R

⁢

I

=

⋃

r

=

0

∞

f

-

r

⁢

I

does not contain the unit element. Let

y

be an

indeterminate

over

K

⁢

[

x

1

,

…

,

x

n

]

, and let

J

be the

inverse image

of

R

⁢

I

under
the

homomorphism

ϕ

:

K

⁢

[

x

1

,

…

,

x

n

,

y

]

→

R

acting as the

identity

on

K

⁢

[

x

1

,

…

,

x

n

]

and sending

y

to

1

/

f

. Then

J

is strictly smaller than

K

⁢

[

x

1

,

…

,

x

n

,

y

]

, so
the

weak Nullstellensatz

gives us an

element

(

a

1

,

…

,

a

n

,

b

)

∈

K

n

+

1

such that

g

⁢

(

a

1

,

…

,

a

n

,

b

)

=

0

for all

g

∈

J

. In
particular, we see that

g

⁢

(

a

1

,

…

,

a

n

)

=

0

for all

g

∈

I

. Our

assumption

on

f

therefore implies

f

⁢

(

a

1

,

…

,

a

n

)

=

0

. However,

J

also contains the element

1

-

y

⁢

f

since

ϕ

sends this element
to zero. This leads to the following

contradiction

:

0

=

(

1

-

y

⁢

f

)

⁢

(

a

1

,

…

,

a

n

,

b

)

=

1

-

b

⁢

f

⁢

(

a

1

,

…

,

a

n

)

=

1

.

The assumption that

f

r

∉

I

for all

r

>

0

is therefore false,
i.e. there is an

r

>

0

with

f

r

∈

I

.

Title

proof of Hilbert’s Nullstellensatz

Canonical name

ProofOfHilbertsNullstellensatz

Date of creation

2013-03-22 15:27:46

Last modified on

2013-03-22 15:27:46

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

4

Author

pbruin (1001)

Entry type

Proof

Classification

msc 13A10

Generated on Fri Feb 9 17:20:34 2018 by

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radical of an integer

radical of an integer

Given a

natural number

n

, let

n

=

p

1

α

1

⁢

⋯

⁢

p

k

α

k

be its

unique factorization

as a

product

of distinct prime powers. Define the of

n

, denoted

rad

⁢

(

n

)

, to be the product

p

1

⁢

⋯

⁢

p

k

. The

radical

of a square-free number is itself.

Title

radical of an integer

Canonical name

RadicalOfAnInteger

Date of creation

2013-03-22 11:45:21

Last modified on

2013-03-22 11:45:21

Owner

KimJ (5)

Last modified by

KimJ (5)

Numerical id

12

Author

KimJ (5)

Entry type

Definition

Classification

msc 13A10

Classification

msc 81-00

Classification

msc 18-00

Synonym

square-free part

Related topic

RadicalOfAnIdeal

Related topic

PowerOfAnInteger

Defines

radical

Generated on Fri Feb 9 17:20:43 2018 by

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unity plus nilpotent is unit

unity plus nilpotent is unit

Theorem

.

If

x

is a

nilpotent element

of a

ring with unity

1 (which may be 0), then the sum

1

+

x

is a unit of the ring.

Proof.

If

x

=

0

, then

1

+

x

=

1

, which is a unit. Thus, we may assume that

x

≠

0

.

Since

x

is

nilpotent

, there is a

positive integer

n

such that

x

n

=

0

. We multiply

1

+

x

by another ring

element

:

(

1

+

x

)

⋅

∑

j

=

0

n

-

1

(

-

1

)

j

⁢

x

j

=

∑

j

=

0

n

-

1

(

-

1

)

j

⁢

x

j

+

∑

k

=

0

n

-

1

(

-

1

)

k

⁢

x

k

+

1

=

∑

j

=

0

n

-

1

(

-

1

)

j

⁢

x

j

-

∑

k

=

1

n

(

-

1

)

k

⁢

x

k

=

1

+

∑

j

=

1

n

-

1

(

-

1

)

j

⁢

x

j

-

∑

k

=

1

n

-

1

(

-

1

)

k

⁢

x

k

-

(

-

1

)

n

⁢

x

n

=

1

+

0

+

0

=

1

(Note that the

summations

include the term

(

-

1

)

0

⁢

x

0

, which is why

x

=

0

is excluded from this case.)

The reversed

multiplication

gives the same result. Therefore,

1

+

x

has a

multiplicative inverse

and thus is a unit.
∎

Note that there is a this proof and geometric series: The goal was to produce a multiplicative inverse of

1

+

x

, and geometric series yields that

1

1

+

x

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

n

,

provided that the summation

converges

(

http://planetmath.org/AbsoluteConvergence

). Since

x

is nilpotent, the summation has a finite number of nonzero terms and thus .

Title

unity plus nilpotent is unit

Canonical name

UnityPlusNilpotentIsUnit

Date of creation

2013-03-22 15:11:54

Last modified on

2013-03-22 15:11:54

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

21

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 13A10

Classification

msc 16U60

Related topic

DivisibilityInRings

Generated on Fri Feb 9 17:20:48 2018 by

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