305 -Rings, ideals, homomorphisms (3)

In order to understand the construction of the quotient ring from last lecture, it is convenient to examine some examples in details. We are interested in ideals {I} of {{\mathbb F}[x],} where {{\mathbb F}} is a field. We write {{\mathbb F}[x]/I} for the quotient ring, i.e., the set of equivalence classes {[a]_\sim} of polynomials {a} in {F[x]} under the equivalence relation {a\sim b} iff {a-b\in I.}

 

  • If {I=\{0\},} then for any {a,} the equivalence class {[a]_\sim} is just the singleton {\{a\}} and the homomorphism map {h:{\mathbb F}[x]\rightarrow{\mathbb F}[x]/I} given by {h(a)=[a]_\sim} is an isomorphism.

To understand general ideals better the following notions are useful; I restrict to commutative rings with identity although they make sense in other contexts as well:

Definition 1 Let {R} be a commutative ring with identity. An ideal {I} is principal iff it is the ideal generated by an element {a} of {R,} i.e., it is the set {(a)} of all products {ab} for {b\in R.}

 

For example, {\{0\}=(0)} is principal. In {{\mathbb Z}} every subring is an ideal and is principal, since all subrings of {{\mathbb Z}} are of the form {n{\mathbb Z}=(n)} for some integer {n.}

Definition 2 An integral domain is a commutative ring {R} with identity wher {1\ne0} and without zero divisors. Recall that {a} is a zero divisor iff {a\ne0} and there is some {b\ne0} such that {ab=0.}

 

For example, in {{\mathbb Z}_n} any nonzero {a=[k]_n} with {(k,n)>1} is a zero divisor: Let {m=(k,n)} and consider {b=[n/m]_n.} Then {ab=0.}

Definition 3 A unit in a commutative ring {R} with identity is an {a} for which there is some {b} such that {ab=1.} We write {R^*} for the set of units of {R.}

 

For example, in {{\mathbb Z}_n,} any {a=[k]_n} with {(n,k)=1} is a unit: Recall that if {(n,k)=1} then there are integers {l,m} such that {nm+kl=1,} so if {b=[l]_n,} then {ab=1.}

If {{\mathbb F}} is a field, {{\mathbb F}^*={\mathbb F}\setminus\{0\},} since every nonzero element has an inverse.

If {{\mathbb F}} is a field, then {{\mathbb F}[x]^*={\mathbb F}^*.} To see this, recall that the degree {{\rm deg}(p(x))} of a polynomial {p(x)} is {-\infty} if {p(x)=0} and it is {n\ge0} if the largest power of {x} that appears in {p} with a nonzero coefficient of {p} is {x^n.} Suppose that {p,q} are nonzero polynomials in {{\mathbb F}[x].} Say that {{\rm deg}(p(x))=n} and {{\rm deg}(q(x))=m,} so there are coeffiecients {a_0,\dots,a_n} and {b_0,\dots,b_m} with {a_n,b_m} not zero, such that

\displaystyle  p(x)=a_0+a_1x+\dots+a_nx^n,\mbox{ and }q(x)=b_0+b_1x+\dots+b_mx^m.

Then {{\rm deg}(p(x)q(x))=n+m,} since the coefficient of {x^k} in the product {pq} is the sum of all the coefficients {a_ib_j} with {i+j=k.} Of course, this is nonzero only if {i\le n} and {j\le m,} so {k\le n+m.} Moreover, the coefficient of {x^{n+m}} in {pq} is {a_nb_m} which is nonzero since a field has no zero divisors.

This shows that {{\rm deg}(pq)={\rm deg}(p)+{\rm deg}(q)} if both {p,q} are nonzero. If one of them is zero, then {pq=0} as well, and {{\rm deg}(pq)={\rm deg}(p)+{\rm deg}(q)} is true due to our convention that {{\rm deg}(0)=-\infty.}

It is clear now that if {p} is a unit in {{\mathbb F}[x],} then {{\rm deg}(p)=0} since otherwise {{\rm deg}(pq)\ge{\rm deg}(p)>0} for any nonzero {q,} but then {pq\ne1.} But {{\rm deg}(p)=0} means that {p} is a nonzero constant, i.e., a nonzero element of {{\mathbb F}.}

Notice that this argument also shows that {{\mathbb F}[x]} is an integral domain, as we showed that {pq\ne0} whenever both {p,q} are nonzero.

Definition 4 An integral domain {R} is a principal ideal domain (pid) iff every ideal of {R} is principal.

 

For example, any field is a principal ideal domain, although this example is somewhat trivial:

More generally, suppose {R} is a commutative ring with identity and let {I} be an ideal of {R.} Suppose that {I} contains a unit. Then {I=R,} because if {a\in I} is a unit then there is some {b\in R} such that {ab=1} and so {a(bc)=c} for any {c\in R.} Since ideals are closed under multiplication by any element of {R,} it follows that {c\in I.} Since {c} was arbitrary, {R=I.}

Now, if {{\mathbb F}} is a field, then every nonzero element of {{\mathbb F}} is a unit, and we have that any ideal is either {(0)} or {{\mathbb F}=(1).}

A more interesting example if {{\mathbb F}[x].} That this is a pid is shown in detail in Chapter 8 of the book, that I highly recommend you study carefully. The argument resembles strongly results we shoed for the integers.

The idea is this:

  • First, in {{\mathbb F}[x]} we have a division algorithm,

so given any nonzero polynomials {a,b} in {{\mathbb F}[x]} there are unique polynomials {q,r} in {{\mathbb F}[x]} with {{\rm deg}(r)<{\rm deg}(b)} and such that {a=bq+r.} Note in particular that if {{\rm deg}(b)=0,} i.e., if {b} is a (nonzero) constant, then {b|a,} as {r} must be zero.

In fact, we can generalize to {{\mathbb F}[x]} the notion of greatest common divisor:

Definition 5 If {a,b} are nonzero polynomials in {{\mathbb F}[x],} a greatest common divisor (gcd) of {a,b} is any nonzero polynomial {r} such that:

  1. {r|a} and {r|b.}
  2. Whenever {s} is a polynomial in {{\mathbb F}[x]} and {s|a} and {s|b,} then {s|r.}

 

So any nonzero polynomials {a,b} may admit more than one gcd, since if {r} is a gcd of {a,b,} then so is {\alpha r} for any {\alpha\in{\mathbb F}^*.} On the other hand, this is the only obstacle to uniqueness: If {r,s} are both gcds of {a,b} then {r|s} and {s|r} so {{\rm deg}(r)={\rm deg}(s)} and there is a unit {\alpha} such that {r=\alpha s.}

 

  • Second, in {{\mathbb F}[x]} we have an Euclidean algorithm,

so by repeated application of the division algorithm we can find a gcd of any two nonzero polynomials {p,q\in{\mathbb F}[x];} moreover, there are polynomials {a,b\in{\mathbb F}[x]} such that {pa+qb} is a gcd of {p,q.}

Using this one can easily show that any ideal in {{\mathbb F}[x]} is principal, just as we did for {{\mathbb Z}.} We’ll revisit the details after the break, and they are in Chapter 8 of the book, but it is important that you notice that we are basically repeating the proofs we already know.

Just as with {{\mathbb N},} once we have a notion of gcd and an Euclidean algorithm, we can talk about prime or irreducible elements in {{\mathbb F}[x].} Once we have this notion, we will revisit the construction of the quotient rings {{\mathbb F}[x]/I} and study their properties.

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One Response to 305 -Rings, ideals, homomorphisms (3)

  1. […] Let’s conclude the discussion from last lecture. […]

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