# Examples of prime ideals that are not maximal

I would like to know of some examples of a prime ideal that is not maximal in some commutative ring with unity.

• sub-rings of $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ and sub-rings in $\mathbb{Z}_n$, but I never thought to check $\{0\}$ – tmpys Feb 28 '14 at 9:15
• Would still love some more interesting ones.... – tmpys Feb 28 '14 at 9:15
• yes I think $k \mathbb{Z}$ for $k\in\mathbb{Z}$ is an ideal. Am I wrong? I am confused by your question;why would I say it if I did not think it...If you can't help, please just don't comment. Or, in the vein of your question,do you think you are helping with your questions? – tmpys Feb 28 '14 at 9:38
• @tmpys You do appear to be confused about the difference between subrings and ideals, and surely it is not unhelpful to point this out to you? (Although I agree that the sentence "I can't confusing you more than you are" is confusing.) – Derek Holt Feb 28 '14 at 10:27
• Another example is $(x^2-y)$ in $\mathbb C[x,y]$, see here. – Martin Sleziak Apr 24 '16 at 5:20

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.

• this is how I convinced myself that there were such things, but I still cant think of good clean examples. – tmpys Feb 28 '14 at 9:20
• @tmpys: You're going to have to use example rings which contain elements that are neither zero divisors nor units (to make it possible to discriminate between prime and maximal ideals). How many such rings do you know? – Eric Towers Feb 28 '14 at 9:30
• Note here that $R$ could itself be a suitable polynomial ring, say $K[y]$ for some suitable $K$. – Mark Bennet Feb 28 '14 at 9:53
• Another concrete way to see why (x) is not maximal is by giving an ideal strictly in between (x) and R[x], like (2,x) which contains all polynomials which have constant term even. – Pranav Bisht Jan 1 '17 at 11:58

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.

• The same example works if $\mathbb Z$ is replaced by any other integral domain, which is not a field. – Martin Sleziak Apr 10 '14 at 7:12

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.

• which is an example of a proper ideal that properly contains $⟨x^2+1⟩$? – Ponce Feb 14 at 0:00

$$\Bbb Z \times \{0\}$$ is a prime ideal of $$\Bbb Z \times \Bbb Z$$, but it is not maximal. It is contained in the proper ideal $$\Bbb Z \times pZ$$.

• This is not a prime ideal. A prime ideal of $R \times S$ must either be $p \times S$ for some prime ideal $p$ of $R$ or $R \times q$ for some prime ideal $q$ of $S$. – Geoffrey Trang Aug 12 '19 at 18:47
• Yes, otherwise the quotient will not be a domain. – TMO Aug 12 '19 at 19:30
• @GeoffreyTrang thanks. I will try to think of another example then. – Justin Meiners Aug 12 '19 at 20:08
• @GeoffreyTrang how is that? – Justin Meiners Aug 12 '19 at 20:24
• This time, your answer works. In general, the cartesian product of ideals from 2 or more rings is an ideal of the product ring, but it will not be prime unless all but one of the factor ideals is equal to the whole ring, and the remaining factor is prime. – Geoffrey Trang Aug 12 '19 at 20:28

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 $X_1^2+X_2^2-1$.

• By the Nullstellensatz ? Let $f_i \in F[X_1,\ldots, X_n]$ some polynomials and $C = \{ a \in F^n, \forall i, f_i(a) = 0\}$ then the Nullstellensatz says the ideal $I = \{ u \in F[X_1,\ldots, X_n], \forall a \in A, u(a) = 0\}$ is the radical of some ideal generated by the $f_i$. – reuns Aug 12 '17 at 10:07
• @reuns The corollary is that maximal ideals are of the form $(X_1-a_1,\dots,X_n-a_n)$. – egreg Aug 12 '17 at 10:17

We can give several examples. The key facts are:

1. An ideal $$I$$ of $$R$$ is prime if and only if $$R/I$$ is an integral domain.
2. 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)

1. $$R = Q[x,y$$] and $$I = (x)$$
2. $$R = Z[x]$$ and $$I = (x)$$
3. $$R = Z$$ and $$I = (0)$$
• Are $Z$ and $Q$ the integers and rational numbers (respectively)? If so, the usual notation is to use \Bbb Z (i.e. $\Bbb Z$) and \Bbb Q (i.e. $\Bbb Q$) for these sets. – Xander Henderson Feb 21 '19 at 17:07