## 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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