Section 27 Prime and Maximal Ideals

27.1 Example

According to Section 19 Integral Domains > 19.11 Theorem ${\mathbb{Z}}_p$ is a field, if p is a prime. ${\mathbb{Z}}_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$. Thus a factor ring of an integral domain may be a field.

27.3 Example

The subset $N=\{0,3\}$ of ${\mathbb{Z}}_6$ (not a integral domain) is easily seen to be an ideal of ${\mathbb{Z}}_6$ ${\mathbb{Z}}_6/N\subseteq {\mathbb{Z}}_3$ has three elements: $0+N$, $1+N$, $2+N$

Remember Section 19 Integral Domains > 19.5 Definition Integral Domain and Section 19 Integral Domains > 19.11 Theorem

NOTE

if R is not even an integral domain, that is, if R has zero divisors, it is still possible for $R/N$ to be a field.

Definition: improper ideal and trivial ideal

A First Course in Abstract Algebra, p.246

Every nonzero ring $R$ has at least two ideals, the improper ideal $R$ and the trivial ideal$\{0\}$

27.5 Theorem

A First Course in Abstract Algebra, p.246

If R is a ring with unity, and N is an ideal (Definition of Ideal) of R containing a unit (Definition of unit), then N = R

Or we can say: no element in N can have a multiplicative inverse Proof: Let $u\in N$ for an unit in $N$ $rN\subseteq N$ with $r\in R$ implies that if we take $r=u^{-1}$, we find that $1$ must be inside $N$ If $1\in N$, then $r1\subseteq N$ and $r1=r$, meaning that every element in $R$ is in $N$.

27.6 Corollary

A First Course in Abstract Algebra, p.246

A field contains no proper nontrivial ideals.

27.7 Definition: maximal ideal

A First Course in Abstract Algebra, p.247

A maximal ideal of a ring R is an ideal M different from R such that there is no proper ideal N of R properly containing M.

So we can also say (just like the maximal normal subgroup in group theory):

NOTE

$M$ is a maximal ideal of $R$ if and only if $R/M$ is a field.

27.10 Example

p is a prime Since $\mathbb{Z}/p\mathbb{Z}$ is isomorphic to ${\mathbb{Z}}_p$ (a field, according to Section 19 Integral Domains > 19.11 Theorem). We see that the maximal ideals of $Z$ are precisely the ideals $p\mathbb{Z}$

Tip

Most rings have multiple maximal ideals. Only local rings have exactly one maximal idea.

27.11 Corollary

A First Course in Abstract Algebra, p.247

A commutative ring with unity is a field if and only if it has no proper nontrivial ideals.

Extension to 27.6 Corollary Proof: If a commutative ring R with unity has no proper nontrivial ideals, then {0} is a maximal ideal and R/{0}, which is isomorphic to R, is a field by the Note in 27.7 Definition maximal ideal

Factor Ring and Integral Domain

The factor ring $R/N$ will be an integral domain if and only if: $(a+N)(b+N)=N⟹a+N=Norb+N=N$ Note that $a+N$ and $b+N$ are two elements in the factor ring $R/N$. This is the same as the definition of an integral domain (no 0 divisor Definition of Integral Domain) $ab=0⟹a=0orb=0$

Remember:

• ${\mathbb{Z}}_n$ is an integral domain if and only if $n$ is prime
• If $p$ is a prime, ${\mathbb{Z}}_p$ is a field

27.13 Definition: prime ideal

A First Course in Abstract Algebra, p.248

An ideal $N\ne R$ in a commutative ring $R$ is a prime ideal if: $ab\in N$ implies that either $a\in N$ or $b\in N$ for $a,b\in R$.

So we cannot have two elements outside of $N$, whose product is in the ideal $N$. And of course $\{0\}$ is a prime ideal in any integral domain, because of no 0 divisor.

Example: $\mathbb{Z}\times \{0\}$ is a prime ideal of $\mathbb{Z}\times \mathbb{Z}$

Characteristics of a commutative ring R with unity

For a commutative ring $R$ with unity:

1. An ideal $M$ of $R$ is maximal if and only if $R/M$ is a field.
2. An ideal $N$ of $R$ is prime if and only if $R/N$ is an integral domain.
3. Every maximal ideal of $R$ is a prime ideal.

27.17 Theorem

A First Course in Abstract Algebra, p.249

If R is a ring with unity 1, then the map $\phi :Z\to R$ given by $\phi (n)=n\cdot 1$ for $n\in Z$ is a homomorphism of $Z$ into $R$.

27.20 Definition Prime Fields

A First Course in Abstract Algebra, p.250

The fields $Z_p$ and $Q$ are prime fields.

27.21 Definition Principal Ideal

In a ring with unity $R$, the ideal $\{ra|r\in R\}$ is the principal ideal generated by a, denoted as $⟨a⟩$.

• An ideal is a principal ideal if such a exists.

Example: Considering the ring $\mathbb{Z}$, the ideals $n\mathbb{Z}$ are generated by n (Note that every ideal of $\mathbb{Z}$ is of the form $n\mathbb{Z}$)

27.24 Theorem: Ideal in polynomial Field

A First Course in Abstract Algebra, p.250

If $F$ is a field, every ideal in $F[x]$ is principal.

Proof: Let $N$ be an ideal of field $F[x]$

• If $N=\{0\}$, then it’s trivial, $\{0\}$ is generated be $⟨0⟩$
• If $N\ne \{0\}$ and the degree of a minimal degree element $g(x)$ is 0 (in other words g(x) is constant), then $N=F[x]=⟨1⟩$ according to 27.5 Theorem, because elements in $N$ are units.

Question

Why is constant g(x) here an unit?

• ⏫ Answer: If $N\ne \{0\}$ and the degree of a minimal degree element $g(x)$ is greater than 0, then:
• consider $f(x)=g(x)q(x)+r(x)$:
• $f(x)\in N$ and $g(x)\in N$, then we have $f(x)-g(x)q(x)=r(x)\in N$ according to the definition of an ideal.
• and since $g(x)$ is a nonzero element of minimal degree in N, we must have $r(x)=0$
• Thus, every $f(x)=g(x)q(x)$, meaning that $N=⟨g(x)⟩$

Tip

In $F[x]$, $f(x)=g(x)q(x)+r(x)$. then $r(x)$ is either 0 or (degree r(x)) < (degree g(x))

27.25 Theorem

A First Course in Abstract Algebra, p.251

An ideal $⟨p(x)⟩\ne \{0\}$ of $F[x]$ is maximal if and only if $p(x)$ is irreducible over F.

irreducible: cannot be fatorized

Proof:

• Let $p(x)=f(x)g(x)$
• $⟨p(x)⟩$ is a maximal ideal $⟶$ $⟨p(x)⟩$ is a prime ideal (see Characteristics of a commutative ring R with unity)
• either $f(x)\in ⟨p(x)⟩$ or $g(x)\in ⟨p(x)⟩$
• meaning that either $f(x)$ or $g(x)$ has $p(x)$ as a factor.
• But then the degrees of $f(x)$ and $g(x)$ cannot both be less than the degree of $p(x)$.
• p is irreducible over F Proof conversely:
• proof⏫

Example: Show that $f(x)=x^3+3x+2$ in ${\mathbb{Z}}_5[x]$ is irreducible over ${\mathbb{Z}}_5$.

• We need to firstly find a term in form $x-a$ for some $a\in {\mathbb{Z}}_5$, so that $f(a)=0$.
• Iterate through $f(a)$ for $a\in {\mathbb{Z}}_5$, we find that such a doesn’t exist, showing that $f(x)$ has no zeros in ${\mathbb{Z}}_5$.
• $f(x)$ is irreducible over ${\mathbb{Z}}_5$
• $⟨x^3+3x+2⟩$ is a maximal ideal
• ${\mathbb{Z}}_5[x]/⟨x^3+3x+2⟩$ is a field (see 27.7 Definition maximal ideal)

Similarly, $x^2-2$ is irreducible in $\mathbb{Q}[x]$, so $\mathbb{Q}[x]/⟨x^2-2⟩$ is a field.

27.27 Theorem

A First Course in Abstract Algebra, p.251

Let p(x) be an irreducible polynomial in F[x]. If p(x) divides r(x)s(x) for r(x), s(x) ∈ F[x], then either p(x) divides r(x) or p(x) divides s(x).

Proof:

• Suppose p(x) is irreducible and divides r(x)s(x)
• $⟨p(x)⟩$ is a maximal ideal (see 27.25 Theorem)
• $⟨p(x)⟩$ is a prime ideal (see Characteristics of a commutative ring R with unity)
• $r(x)s(x)\in ⟨p(x)⟩$ implies either $r(x)\in ⟨p(x)⟩$ or $s(x)\in ⟨p(x)⟩$ (see 27.13 Definition prime ideal)
• Hence, $p(x)$ divides $r(x)$ or $p(x)$ divides $s(x)$ (see 27.21 Definition Principal Ideal) $q.e.d.$