Prime and maximal ideals

Equivalence. Proof.

Prime

Proof. Forward & reverse. $R/I$ is an integral domain iff:

$$\mathrm{\forall }c,d\in R/I,cd=0⟹(c=0\text{ or }d=0)$$

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

$$\mathrm{\forall }a,b\in R,ab\in I⟹(a\in I\text{ or }b\in I)$$

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:

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 $⟹$ a field. And so an ideal $I$ that induces a finite $R/I$ quotient, $I$ is prime $⟺$ 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:

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