I would like to know of some examples of a prime ideal that is not maximal in some commutative ring with unity.
$\endgroup$ 56 Answers
$\begingroup$Let $R$ be an integral domain and consider $R[x]/(x) \cong R$. It's not a field (unless $R$ is), so $(x)$ is not maximal. Since $R$ has no zero divisors, $(x)$ is a prime ideal.
$\endgroup$ 4 $\begingroup$Take $(0)$, the zero ideal, in $\mathbb{Z}$, which is prime as the integers are an integral domain, but not maximal as it is contained in any other ideal.
$\endgroup$ 1 $\begingroup$A concrete example: $\langle x^2+1\rangle$ is a prime ideal in $\mathbb{Z}[x]$, but is not maximal, since $\mathbb{Z}[x]/\langle x^2+1\rangle$ is isomorphic to the Gaussian integers $\mathbb{Z}[i]$, which is an integral domain that is not a field.
$\endgroup$ 2 $\begingroup$$\Bbb Z \times \{0\} $ is a prime ideal of $\Bbb Z \times \Bbb Z$, but it is not maximal. It is contained in the proper ideal $\Bbb Z \times pZ$.
$\endgroup$ 5 $\begingroup$There is plenty of interesting examples!
Let $F$ be an algebraically closed field and consider an irreducible polynomial $f(X_1,\dots,X_n)$ in the polynomial ring $F[X_1,\dots,X_n]$. Then the ideal $I$ generated by $f$ is prime and not maximal, by Hilbert’s Nullstellensatz.
For the case $n=2$ and $F=\mathbb{C}$ the geometric interpretation is an irreducible curve, for instance the parabola defined by $X_1^2-X_2$ or the circle defined by $X_1^2+X_2^2-1$.
$\endgroup$ 2 $\begingroup$We can give several examples. The key facts are:
- An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
- An ideal $I$ of $R$ is maximal if and only if $R/I$ is a field.
Various examples (check yourself these examples satisfies your condition)
- $R = Q[x,y$] and $I = (x)$
- $R = Z[x]$ and $I = (x)$
- $R = Z$ and $I = (0)$
