Algebra I - Commutative Rings
Definitionen und Sätze für moderne Algebra
Definitionen und Sätze für moderne Algebra
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Cartes-fiches
Def: (Ring)
A ring R is a set with two binary operations RxR -> R: addition (a,b) -> a+b and multiplication (a,b)->ab sucht that
- R is an abelian group under addition
- Associativity
- Neutral element for multiplication
- Distributivity
Def: (Subring)
A subset S of a ring R is a subring of R if
- 1 is element of S
- if a,b are in S, then a+b is in S
- if a,b are in S, then ab is in S
Def: (Commutative Ring)
A ring R is commutative, if ab = ba for all a and b in R
Def: (Domain)
A domain (or integral domain) is a commutative ring R that satisifies two extra axioms:
- 1 is not equal to zero
- Cancellation Law (for all a,b in R: if ca = cb and c not zero then a = b)
Question: When is \(\mathbb{Z}_m\) a domain?
\(\mathbb{Z}_m\) is a domain, if and only of m is prime
Def: (Unit)
An element u in a commutative Ring R is called a unit, if u | 1 in R, that is if there exists an v in R with uv = 1
The element v is called the multiplicative inverse of u and denotetd with u-1
Def: (Field)
A field F is a commutative ring in which \(1 \neq 0\) and every nonzero element a is a unit. that is, there is a a-1 in F with a-1a=1.
Question: Is it true, that every field F is a domain?
Theorem: If R is a domain, then there is a field containing R as a subring
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Def: (Fraction Field)
The field F constructed from R (the set of all equivalence classes of (a,b) by (a,b) = (c,d) if ad = bc) is called the fractionfield of R, denoted by: Frac(R)
Theorem: (Unique Factorization)
if k is a field, then every polynomial f(x) in k[x] of degree >= 1 is a product of a nonzero constant and monic irreducibles and
\(f(x) = ap_1(x) \dots p_m(x)\)
this factorization is unique bis auf eine umsortierung der Reihenfolge.....
Def: (irreducible element of a domain)
An element p in a domain R is irreducible if p is not zero and p is no unitg and there exist no none units u,v such that
p = uv
Theorem: If k is a field, then every ideal I in k[x] is a principal ideal
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Def: (principal ideal)
Zu deutsch: Hauptideal
In the mathematical field of ring theory, a principal ideal is an ideal I in a ring R that is generated by a single element a of R through multiplication by every element of R.
Theorem: (Divison Algorithm)
If k is a field an f,g in k[x] with f not zero, then there are unigque polynomials q and r sucht that:
g = qf + r
wehtere either r = 0 or deg(r) < deg(f)
Def: (indeterminate x in R[x])
The indeterminate x in R[x] is
\(x = (0,1,0,0,\dots)\)
Theorem: Let R be a commutative ring and let \(\sigma,\tau \in R[x]\) be nonzereo polynomials.
- \(deg(\sigma\tau) \leq deg(\sigma) \cdot deg(\tau)\)
If R is a domain, then \(\sigma \tau \neq 0\) and
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Def: (addition form formal power series and multiplication from formal power series)
\(\sigma + \tau = (s_0 + t_0, s_1 + t_1, \dots , s_n + t_n, \dots)\)
\(\sigma \tau = (c_0, c_1, \dots)\)
with \(c_k = \sum_{i = 0}^k s_i\cdot t_{k-i}\) wehre k = i+j
Def: (poliynomial)
A polynomial over a commutative ring R is a formal power series
\(\sigma = (s_0,s_1,\dots,s_n,0,\dots,0,\dots)\)
over R for which there exists some integer \(n \geq0\) with \(s_i = 0 \text{ for all} i > n\)
So a polynomial has only finetly many nonzero coefficients.
We call \(s_n\) the leading coefficient of \(\sigma \) and we denote the degree by \(deg(\sigma) = n\)
Def: (formal power series)
If R is a commutative ring, then a formal power series over R is a sequence of elements \(s_i \in R \text{ for all } i\geq 0\) called the coefficients of \(\sigma \):
\(\sigma = (s_0, s_1, \dots ,s_i, \dots)\)
Def: (Prime Ideal)
Let R be a commutative ring an I a subset of R which is a ideal
I is prime ideal if and only of R/I is a integral domain
Def: (Maximal Ideal)
Let R be a commutative ring and I a subset of R which is an ideal.
I is maximal if and only if R/I is a field
Theorem: (First Isomorphism Theorem)
Let R and A be commutative rings. If \(\phi: R \rightarrow A \) is a homomorphism, then \(ker(\phi)\) is an deal in R and \(im(\phi)\) is a subring of A and it satisfy
\(R / ker(\phi) \cong im(\phi)\)
Def: (natural map / projection)
Let I be an ideal in a commutative ring R. The natural map or projection is the function:
\(\pi: R \rightarrow R/I, x \rightarrow [x]\)
Theorem: If I is an ideal in a commutative Ring R, then R/I is a commutative ring.
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Def: (Quotient Ring of R modulo I)
The ring constructed trough the "Congruence Mod I" equivalence relation is called the Quotient Ring of R modulo I and is denoted by: R/I, so it is the set of all equivalent classes.......
Def: (Congruence Mod I)
Given is an ideal I in a commutative Ring R. The relation ~on R, defined by:
\(a \sim b \Leftrightarrow a-b \in I\)
and it is a equivalence relation on R
Def: (Ideal generated by \((b_1, b_2, \dots , b_n)\))
If \(b_1,b_2,\dots,b_n\) lie in R, then the set of all linear combinations:
\(I = \{r_1b_1 + r_2b_2 + \dots + r_nb_n \quad | \quad r_i \in R \quad \forall i\}\)
is an ideal in R an we write \((b_1, \dots , b_n)R\)
Prop: An ring homomorphism is injective if and only if his kern is trivial
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Theorem : If \(\phi: A \rightarrow R\) is a homomorphism, then \(ker(\phi)\) is an ideal in A and \(im(\phi)\) is a subring of R
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Def: (ideal)
An ideal in a commutative ring R is a subset I of R such that :
\(0 \in I\)
\(a,b \in I \Rightarrow a+b \in I\)
\(a \in I, b\in R \Rightarrow ab \in I\)
Def: (kernel and image of a ring homomorphism)
\(ker(\phi) := \{a \in A \quad | \quad \phi(a) = 0 \}\)
\(im(\phi) := \{\phi(a) \quad | \quad a \in A \}\)
Def: (Ring Homomorphism)
If A,B are rings, a ring homomorphism is a function \(\phi: A \rightarrow B\) such that:
\(\phi(1) = 1\)
\(\phi(a + b) = \phi(a) + \phi(b), \quad a,b \in A\)