Notes - Rings and Modules HT24, Prime and maximal ideals

• $I \ne R$
• $I$ is not strictly contained in any other proper ideal.

• $I \ne R$
• $\forall a, b \in R$, if $ab \in I$ then $a \in I$ or $b \in I$ (inclusive).

Just $I \ne R$.

For the forward direction, suppose $(a + I)(b + I) = 0 + I$ for some $a, b \in R$. Then it must be the case that $ab \in I$ so since $I$ is prime, $a \in I$ or $b \in I$. Then it follows $a + I = 0 + I$ or $b + I = 0 + I$, so $R/I$ is an integral domain.

On the other hand if we know $(a + I) = 0 + I$ or $(b + I) = 0 + I$, then $a$ or $b$ lies in $I$. Since we also have that $ab \in I$, it follows $I$ is prime.

Suppose $ab \in I$ and $a \notin I$. Then $I \subsetneq I + \langle a \rangle$, so $I + \langle a \rangle = R$ (since $I$ is maximal). Hence $\exists i \in I, r \in R$ such that $i + ra = 1$, hence multiplying through by $b$ we see $bi + rab = b$. Then since $i \in I$ and $ab \in I$, we see $bi \in I$ and $rab \in I$. Hence $b \in I$.

Alternative proof: $I$ is prime iff $R / I$ is an integral domain. But if $I$ is maximal, then $R/I$ is a field, which is in particular an integral domain.

Suppose $R/I$ is a field. Then $R/I$ has no nontrivial ideals, so by the correspondence theorem, there are no ideals in $R$ containing $I$ that aren’t $I$ or $R$ itself, which implies that $I$ is maximal. On the other hand, if $I$ is maximal, then again by the correspondence theorem, the ideals in $R/I$ are trivial, so $R$ is a field.

\[\langle d \rangle \text{ is a prime ideal} \iff d \text{ is irreducible in } R \iff \langle d \rangle \text{ is a maximal ideal in } R\]

• $(1) \implies (2)$: suppose $d = ab$. Since $d$ is prime (it is a generator of a prime ideal), either $a \in \langle d \rangle$ or $b \in \langle d\rangle$. Wlog assume $a \in \langle d \rangle$. $a$ is not a unit since otherwise $R = \langle a \rangle \subseteq \langle d\rangle$. Then $\exists r \in R$ s.t. $a = rd$, so $d = rdb$. Then $d(1 - rb) = 0$, so $rb = 1$ and hence $b$ is a unit. Then $d$ is irreducible.

• $(2) \implies (3)$: we want to examine the relationship between $\langle d \rangle$ and $I$ when $\langle d \rangle \subseteq I \trianglelefteq R$. Since $R$ is a PID, $I = \langle a \rangle$ for some $a \in R$. $\langle d \rangle \subseteq \langle a \rangle$, so $d = ab$ for some $b \in R$. But then either $a$ or $b$ is a unit. If $a$ is a unit, then $\langle a \rangle = R$, which means $\langle d \rangle$ could still be maximal. If $b$ is a unit, $d$ and $a$ are associates, so $\langle d \rangle = \langle a \rangle = I$. So $\langle d \rangle$ is a maximal ideal.

• $(3) \implies (1)$: this is true for general rings (the proof goes like this: suppose $ab \in I$ and $a \notin I$. Then $I \subseteq I + \langle a \rangle$, so $I + \langle a \rangle$. Hence $1 \in I + \langle a \rangle$, so $1 = i + ra$ for some $i \in I$, $r \in R$. But then multiplying through by $b$, we see $bi + rab = b$, and note the LHS is in $I$, so $b \in I$, or you can also use the fact any field is an integral domain and consider the quotient $R/\langle a \rangle$)

Again, for $d \ne 0$:

\[(1)\\; \langle d \rangle \text{ is a prime ideal} \iff (2)\\; d \text{ is irreducible in } R \begin{array}{c} \centernot\implies \\ \impliedby \end{array} (3)\\; \langle d \rangle \text{ is a maximal ideal in } R\]

An example to show that $d$ being irreducible doesn’t guarantee that $\langle d \rangle$ is maximal: $\mathbb Z[x]$ is a UFD and $2$ is irreducible. But $\langle 2\rangle$ is not maximal:

\[\langle 2 \rangle \subsetneq \langle 2, x\rangle \subsetneq \mathbb Z[x]\]

Again, for $d \ne 0$:

\[(1)\\; \langle d \rangle \text{ is a prime ideal} \iff (2)\\; d \text{ is irreducible in } R \begin{array}{c} \centernot\implies \\ \impliedby \end{array} (3)\\; \langle d \rangle \text{ is a maximal ideal in } R\]

An example to show that $d$ being irreducible doesn’t guarantee that $\langle d \rangle$ is maximal: $\mathbb Z[x]$ is a UFD and $2$ is irreducible. But $\langle 2\rangle$ is not maximal:

\[\langle 2 \rangle \subsetneq \langle 2, x\rangle \subsetneq \mathbb Z[x]\]

$\langle p \rangle$ is a prime ideal, or equivalently if $p \mid ab$, then either (inclusive) $p \mid a$ or $p \mid b$.