Proposition 17.1.
Assume \(R\) is a unital ring and \(I\) is an ideal of \(R\text{.}\) The ideal \(I\) is proper if and only if \(I\) does not contain any units.
Chapter 17 Prime and Maximal Ideals
Section 17.1 Proper Ideals
Every ring, \(R\text{,}\) has at least two ideals: \(\{0\}\) and \(R\text{.}\) We call the ideal \(R\) the improper ideal and say every other ideal is a proper ideal. The ideal \(\{0\}\) called the trivial ideal. We would like to know when a ring contains at least one proper, non-trivial ideal.
Proof.
While the given statement is the more useful, it is simpler to prove the equivalent statement: The ideal \(I\) is improper if and only if \(I\) contains a unit. First assume that \(I\) is improper. Then \(I = R\) contains \(1\) because \(R\) was assumed to be unital. Therefore if \(I\) is improper, then \(I\) contains a unit.
Conversely, assume that \(I\) contains a unit, \(u\text{.}\) Since \(I\) is closed under multiplication,
\begin{equation*}
1 = uu^{-1} \in I\text{.}
\end{equation*}
Hence for every \(r \in R\text{,}\) \(r = 1r \in I\text{.}\) Therefore if \(I\) contains a unit, then \(I\) is improper.
Corollary 17.2.
A field, \(F\text{,}\) contains no non-trivial, proper ideals.
Proof.
Assume \(I\) is a non-trivial ideal. Hence there exists \(x \in I \setminus \{0\}\text{.}\) Since \(F\) is a field, \(x\) is a unit. Therefore \(I\) is improper by Proposition 17.1.
Section 17.2 Maximal Ideals
Definition 17.3.
A maximal ideal of a ring \(R\) is a proper ideal that is not contained in any other proper ideal.
Example 17.4.
Example 17.5.
By Example 16.4, the ideals of the ring \(\Z/60\Z\) are in one-to-one correspondence with the subgroups of \(\Z/n\Z\text{.}\) There is a unique subgroup of \(\Z/60\Z\) of order \(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\) by Corollary 7.30 and are enumerated below.
\begin{equation*}
\langle 0 \rangle,
\langle \overline{30} \rangle,
\langle \overline{20} \rangle,
\langle \overline{15} \rangle,
\langle \overline{12} \rangle,
\langle \overline{10} \rangle,
\langle \overline{6} \rangle,
\langle \overline{5} \rangle,
\langle \overline{4} \rangle,
\langle \overline{3} \rangle,
\langle \overline{2} \rangle,
\langle \overline{1} \rangle
\end{equation*}
To find the maximal ideals we first eliminate those proper ideals which are contained inside another proper ideal:
\begin{align*}
\langle\overline{30}\rangle \amp\leq \langle\overline{5}\rangle
\amp \langle\overline{20}\rangle \amp\leq \langle\overline{5}\rangle
\amp \langle\overline{15}\rangle \amp\leq \langle\overline{5}\rangle
\amp \langle\overline{4}\rangle \amp\leq \langle\overline{2}\rangle\\
\langle\overline{12}\rangle \amp\leq \langle\overline{4}\rangle
\amp \langle\overline{10}\rangle \amp\leq \langle\overline{5}\rangle
\amp \langle\overline{6}\rangle \amp\leq \langle\overline{2}\rangle
\end{align*}
The remaining ideals are the ones generated by \(\overline{5}\text{,}\) \(\overline{3}\text{,}\) and \(\overline{2}\text{.}\) Proving that any one of these is maximal is essentially the same argument, so we show that \(I = \langle\overline{5}\rangle\) is maximal.
Assume \(J\) is an ideal that contains \(I\text{.}\) By Example 16.4, the underlying set \(J\) is equal to a subgroup of order \(d\text{,}\) where \(d\) is a divisor of \(60\text{.}\) Now we observe that \(12\) is a divisor of \(d\) by Lagrange’s Theorem. However, the only divisor of \(60\) larger than \(12\) is \(60\text{.}\) Hence \(J = \Z/n\Z\) is an improper ideal. Therefore \(I = \langle\overline{5}\rangle\)
Remark 17.6.
Observe that each of the groups generated by \(\overline{5}\text{,}\) \(\overline{3}\text{,}\) and \(\overline{2}\) have index \(5\text{,}\) \(3\text{,}\) and \(2\) in \(\Z/60\Z\text{,}\) respectively. Since every group of prime order is cyclic, it follows that as abelian groups, the quotients are isomorphic to \(\Z/5\Z\text{,}\) \(\Z/3\Z\text{,}\) and \(\Z/2\Z\text{,}\) respectively. However, recall from Theorem 16.5 that we defined the ring quotient \(R/I\) to be the abelian group equipped with the multiplication
\begin{equation*}
(a + I)(b + I) = (ab) + \text{.}
\end{equation*}
Hence these quotients are isomorphic as rings to the fields \(\Z/5\Z\text{,}\) \(\Z/3\Z\text{,}\) and \(\Z/2\Z\text{,}\) respectively. It turns out this is not a coincidence, but a defining feature of maximal ideals in commutative, unital rings.
Lemma 17.7.
Assume \(R\) is a commutative, unital ring and \(a \in R\text{.}\) The set
\begin{equation*}
Ra := \left\{xa \;\middle\vert\; x \in R\right\}
\end{equation*}
is an ideal of \(R\text{.}\)
Proof.
First we show that \(Ra\) is a subgroup. This set is non-empty because \(a = 1a \in Ra\text{.}\) For all \(ra, sa \in Ra\text{,}\) \(ra - sa = (r - s)a \in Ra\text{.}\) Hence \(Ra \leq R\) by The Subgroup Criterion
Next, we show that \(Ra\) is closed under multiplication on the left. For all \(ra \in Ra\) and \(s \in R\text{,}\) \(s(ra) = (sr)a \in Ra\text{.}\) Therefore \(Ra\) is an ideal.
Definition 17.8. Principal Ideal.
Assume \(R\) is a commutative unital ring. We say an ideal \(I\) of \(R\)is a principal ideal if there exists \(a \in R\) such that \(I = Ra\text{.}\) We frequently call \(Ra\) the (principal) ideal generated by \(a\).
Theorem 17.9.
Assume \(R\) is a commutative, unital ring and \(I\) is an ideal of \(R\text{.}\) The ideal \(I\) is maximal if and only if the ring \(R/I\) is a field.
Proof.
First assume that \(I\) is a maximal ideal. Let \(\pi \colon R \to R/I\) be the canonical projection. Since \(I\) is proper, we observe that \(R/I\) is not the zero ring and thus \(0 + I \neq 1 + I\text{.}\) Let \(a + I \in R/I\) be any non-zero element and note this implies \(a \not \in I\text{.}\) Suppose to the contrary that \(a + I\) is not a unit. Consider the principal ideal \(J = (R/I)(a + I)\text{.}\)
First we prove that \(1 + I \not\in J\text{.}\) Suppose to the contrary that \(1 + I \in J\text{.}\) Then there exists \(b + I \in R/I\) such that
\begin{equation*}
(b + I)(a + I) = 1 + I\text{,}
\end{equation*}
but this contradicts the assumption that \(a + I\) is not a unit. Hence \(1 + I \not \in J\text{.}\)
Now we prove that \(\pi^{-1}(J)\) is a proper ideal that properly contains \(I\text{.}\) To see that \(\pi^{-1}(J)\) is proper, observe that \(\pi(1) = 1 + I \not \in J\) implies \(1 \not \in \pi^{-1}(J)\text{.}\) Next we observe \(I \subseteq \pi^{-1}(J)\) because for every \(x \in I\text{,}\)
\begin{equation*}
\pi(x) = x + I = 0 + I \in J\text{.}
\end{equation*}
Since \(\pi(a) = a + I \in J\) and \(a \not \in I\text{,}\) this containment is strict, contradicting the assumption that \(I\) is maximal. Hence every non-zero element of \(R/I\) is a unit. Therefore if \(I\) is maximal, then \(R/I\) is a field.
Conversely, assume that \(R/I\) is a field. Note that by Definition 13.27, \(0 + I \neq 1 + I\) implies \(I\) is a proper ideal of \(R\) because \(1 \not \in I\text{.}\) Suppose that \(J\) is an ideal of \(R\) that contains \(I\text{.}\) Since \(R/I\) is a field, the ideal \(\pi(J)\) in the field \(\pi(R) = R/I\) is either the trivial or the improper ideal. In the first case, \(J \subseteq \ker{\pi} = I\) implies \(I = J\text{.}\) In the second case, there exists \(a \in J\) such that \(1 + I = a + I \iff 1 - a \in I \subseteq J\text{.}\) Hence \(1 = a + (1 - a) \in J\) implies \(J = R\text{.}\) Therefore if \(R/I\) is a field, then \(I\) is maximal.
Section 17.3 Prime Ideals
Definition 17.10.
A prime ideal of a ring \(R\) is an ideal, \(I\text{,}\) such that for all \(a,b \in R\text{,}\) if \(ab \in I\text{,}\) then \(a \in I\) or \(b \in I\text{.}\)
Example 17.11.
If \(R\) is an integral domain, then the trivial ideal is a prime ideal. This is because \(R\) has no zero divisors, so \(ab = 0\) implies \(a = 0\) or \(b = 0\text{.}\)
Example 17.12.
If \(p\) is a prime number, then the principal ideal \(\Z p = p\Z\) is a prime ideal. Let \(a,b \in \Z\) be given. Assume \(ab \in p\Z\text{;}\) that is, there exists \(k \in \Z\) such that \(ab = kp\text{.}\) Suppose that \(a \not \in p\Z\) or, equivalently, \(p\) does not divide \(a\text{.}\) By Proposition A.9, there exist \(x,y \in \Z\) such that \(ax + py = 1\text{.}\) Hence
\begin{equation*}
b = b(ax + py) = abx + bpy = kpx + bpy = p(kx + by) \in p\Z\text{.}
\end{equation*}
Therefore \(p\Z\) is a prime ideal of \(\Z\text{.}\)
Theorem 17.13.
Assume \(R\) is a commutative unital ring and \(I\) is an ideal of \(R\text{.}\) The ideal \(I\) is prime if and only if \(R/I\) is an integral domain.
Proof.
Assume \(I\) is a prime ideal. Let \(a + I, b + I \in R/I\) be given and assume \((a + I)(b + I) = 0 + I\text{.}\) By definition,
\begin{equation*}
0 + I = (a + I)(b + I) = (ab) + I \iff ab \in I\text{.}
\end{equation*}
Since \(I\) is assumed to be prime, either \(a \in I\) or \(b \in I\text{.}\) Hence \(a + I = 0 + I\) or \(b + I = 0 + I\text{.}\) Therefore if \(I\) is a prime ideal, then \(R/I\) is an integral domain.
Conversely, assume that \(R/I\) is an integral domain. Let \(a,b \in R\) be given and assume \(ab \in I\text{.}\) Then
\begin{equation*}
0 + I = (ab) + I = (a + I)(b + I)
\end{equation*}
implies \(a + I = 0 + I\) or \(b + I = 0 + I\text{.}\) Hence \(a \in I\) or \(b \in I\text{.}\) Therefore if \(R/I\) is an integral domain, then \(I\) is a prime ideal.
Corollary 17.14.
Proof.
Since \(M\) is maximal, \(R/M\) is a field, hence an integral domain. Therefore \(M\) is a prime ideal.
Warning 17.15.
While every maximal ideal is prime, not every prime ideal is maximal.
Example 17.16.
Consider the ring of polynomials with integer coefficients, \(\Z[x]\text{.}\) The ideal
\begin{equation*}
x\Z[x] = \left\{xf(x) \;\middle\vert\; f \in \Z[x]\right\}
\end{equation*}
is a prime ideal that is not maximal because \(\Z[x]/x\Z[x] \cong \Z\text{.}\) We proceed by showing the map
\begin{align*}
\phi \colon \Z[x] \amp\to \Z\\
f \amp\mapsto f(0)
\end{align*}
is a surjective morphism of rings with kernel \(x\Z[x]\)
The function \(\phi\) is a morphism of rings because arithmetic of polynomials agrees with arithmetic of functions; that is, for all \(f,g \in \Z[x]\text{,}\)
\begin{equation*}
\phi(f + g) = (f + g)(0) = f(0) + g(0) = \phi(f) + \phi(g)
\end{equation*}
and
\begin{equation*}
\phi(fg) = (fg)(0) = f(0)g(0) = \phi(f)\phi(g)\text{.}
\end{equation*}
It is surjective because for every \(n \in \Z\text{,}\) the polynomial \(f(x) = n\) satisfies \(\phi(f) = f(0) = n\text{.}\)
Now observe that for any \(f(x) = \sum_{n=0}^\infty a_nx^n \in \Z[x]\)
\begin{align*}
f \in \ker\phi \amp\iff
0 = \phi(f) = f(0) = a_0\\
\amp\iff f(x) = \sum_{n=1}^\infty a_nx^n = x\sum_{n=0}^\infty a_nx^n\\
\amp\iff f \in x\Z[x]
\end{align*}
Therefore
\begin{equation*}
\frac{\Z[x]}{x\Z[x]} = \frac{\Z[x]}{\ker\phi} \cong \phi(\Z[x]) = \Z
\end{equation*}
by the First Isomorphism Theorem.
