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Math and science::Algebra::Aluffi

Prime and maximal ideals

Prime and maximal ideals

Let \( R \) be a commutative ring. Let \( I \neq (1) \) be an ideal of \( R \).

Prime
\( I \) is said to be a prime ideal iff \( R / I \) is an integral domain.
Maximal
\( I \) is said to be a maximal ideal iff \( R / I \) is a field.

The above definition is equivalent to the following conditions.

Let \( R \) be a commutative ring. Let \( I \neq (1) \) be an ideal of \( R \).

Prime
\( I \) is a prime ideal iff \[ \forall a, b \in R, \;\; ab \in I \implies ( (a \in I \text{ or } b \in I) \]
Maximal
\( I \) is a maximal ideal iff \[ \text{for all ideals } J \subseteq R, \;\; I \subseteq J \implies (I = J \text{ or } J = R ) \]

Can you construct the proof of this equivalence?


Equivalence. Proof.

Prime

Proof. Forward & reverse. \( R/I \) is an integral domain iff:

\[ \forall c,d \in R/I, \;\; c d = 0 \implies (c = 0 \text{ or } d = 0) \]

The RHS and LHS of the above is equivalent the RHS and LHS of the following:

\[ \forall a,b \in R, \;\; ab \in I \implies (a \in I \text{ or } b \in I) \]
Maximal

Proof. Forward. No ideal other than \( R \) contains \( I \) implies that \( R / I \) has no non-trivial ideals (\( (0) \) or \( ( (1)) \), which is enough to say it's a field, which makes \( I \) maximal.

Reverse. Being a field, the only ideals of \( R/I \) are the ideals \( (0) \) and \( (1) \). And so in \( R \), there can be no ideals containg \( I \) other than \( R \).

Integer prime analogy

We saw previously that:

\[ \mathbb{Z}/p\mathbb{Z} \text{ an integral domain } \iff \mathbb{Z}/p\mathbb{Z} \text{ a field } \iff p \text{ is prime} \]

Knowing that \( (p) = p\mathbb{Z} \) represent the same ideal in \( \mathbb{Z} \), if we look at a prime \( p \) through the lens of this card, we could say that primes get their "primeness" by inducing an integral domain through the quotient \( \mathbb{Z}/(p) \), with \( (p) \) being a prime ideal. But \( (p) \) is also a maximal ideal, so it is interesting to ask: why we didn't give "prime ideal" to the concept of maximal ideal? Part of the reason might be that finite integral domain \( \implies \) a field. And so an ideal \( I \) that induces a finite \( R/I \) quotient, \( I \) is prime \( \iff \) maximal.

Example

\( (x - a) \)

For a polynomial ring \( R[x] \), \( R[x]/(x-a) \cong R \), and so \( (x-a) \) is prime iff \( R \) is an integral domain, and it is maximal iff \( R \) is a field.

\( (2,x) \)

What about \( R[x]/(2,x) \)? We we have:

\[ \mathbb{Z}[x]/(2,x) \cong \frac{\mathbb{Z}/(x)}{(2)} \cong \mathbb{Z}/(2) \cong \mathbb{Z}/2\mathbb{Z} \]

So the ideal \( (2,x) \) is maximal in \( \mathbb{Z}[x] \). Intuitively, all terms to the left of the constant don't distinguish any equivalence class, and there are only two distinct equivalence classes for the constant terms.

Source

Aluffi p150