Modern Algebra Exam 2

Front 1

Slow p- subgroup

Back 1

of G if|P|= p^α for some integer α≥1 such that p^α|n, but p^α+1|/n.

Front 2

Simple group

Back 2

If it does not contain a proper non-trivial normal subgroup

Front 3

Ring

Back 3

Nonempty set R with two binary operations + and • which satisfies the following properties:

1. (R,+) is an abelian group

2. Multiplication in R is associative

3. The distributive laws hold

Front 4

Ring with unity

Back 4

If it has a multiplicative identity element, usually denoted 1 or 1r

Front 5

Commutative ring

Back 5

Of the multiplication operation in R is commutative

Front 6

Subring

Back 6

if S is a ring under the operations of + and · inherited from R, and S contains the unity element 1R of R.

Front 7

Unit

Back 7

if there exists an element a^−1∈R such that aa^−1 =1.

Front 8

Zero Divisor

Back 8

if there exists a nonzero element b∈R such that ab =0

Front 9

Integral Domain

Back 9

if R does not contain any nonzero zero divisors

Front 10

Field

Back 10

if every nonzero element of R is a unit.

Front 11

Multiplicative Group of Units

Back 11

The set R× ={a∈R : a is a unit

Front 12

Nilpotent

Back 12

if there exists n≥1 such that a^n =0.

Front 13

Nilradical

Back 13

is the set of all nilpotent elements in R. That is, nil(R)={a∈R: a^n =0 for some n≥1}.

Front 14

Ideal

Back 14

A nonempty subset I of R is said to be an ideal of R if

(i) a±b∈I for all a,b∈I (I is closed under addition and subtraction)

(ii) ra∈I for all a∈I and r∈R (I absorbs multiplication from R)

Front 15

Principal Ideal

Back 15

The set (a)=Ra = aR={ra : r∈R}

Front 16

Principal Ideal Domain

Back 16

if every ideal in R is principal

Front 17

Ideal Sum

Back 17

I +J ={x + y : x∈I and y∈J}.

Front 18

Ideal Product

Back 18

IJ = {SUM(n to i=1) aibi : ai ∈I,bi ∈J,n≥1}

Front 19

Radical of an Ideal

Back 19

sqrt(I) ={a∈R: a^n∈I for some n≥1}.

Front 20

Quotient Ring

Back 20

R/I ={a+I : a∈R}

is a commutative ring with unity under the operations

(a+I)+(b+I)=(a+ b)+I

(a+I)(b+I)= ab+I