I would like to know of some examples of a prime ideal that is not maximal in some commutative ring with unity.
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1$\begingroup$ sub-rings of $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ and sub-rings in $\mathbb{Z}_n$, but I never thought to check $\{0\}$ $\endgroup$– tmpysFeb 28, 2014 at 9:15
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$\begingroup$ Would still love some more interesting ones.... $\endgroup$– tmpysFeb 28, 2014 at 9:15
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$\begingroup$ 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? $\endgroup$– tmpysFeb 28, 2014 at 9:38
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$\begingroup$ @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.) $\endgroup$– Derek HoltFeb 28, 2014 at 10:27
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$\begingroup$ Another example is $(x^2-y)$ in $\mathbb C[x,y]$, see here. $\endgroup$– Martin SleziakApr 24, 2016 at 5:20
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6 Answers
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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.
answered Feb 28, 2014 at 9:12
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1$\begingroup$ this is how I convinced myself that there were such things, but I still cant think of good clean examples. $\endgroup$– tmpysFeb 28, 2014 at 9:20
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1$\begingroup$ @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? $\endgroup$ Feb 28, 2014 at 9:30
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1$\begingroup$ Note here that $R$ could itself be a suitable polynomial ring, say $K[y]$ for some suitable $K$. $\endgroup$ Feb 28, 2014 at 9:53
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4$\begingroup$ 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. $\endgroup$ Jan 1, 2017 at 11:58
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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.
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5$\begingroup$ The same example works if $\mathbb Z$ is replaced by any other integral domain, which is not a field. $\endgroup$ Apr 10, 2014 at 7:12
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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.
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1$\begingroup$ which is an example of a proper ideal that properly contains $⟨x^2+1⟩$? $\endgroup$– PonceFeb 14, 2021 at 0:00
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$\begingroup$ @zjs $(2, x^2+1)=(2, (x+1)(x-1))$, which is not prime. $(p, x^2+1)$ for any prime $p\cong_4 3$ works. $\endgroup$ Oct 9 at 19:33
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$\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 $\Bbb Z \times pZ$.
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2$\begingroup$ 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 $S$. $\endgroup$ Aug 12, 2019 at 18:47
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$\begingroup$ Yes, otherwise the quotient will not be a domain. $\endgroup$– ConAug 12, 2019 at 19:30
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$\begingroup$ @GeoffreyTrang thanks. I will try to think of another example then. $\endgroup$ Aug 12, 2019 at 20:08
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3$\begingroup$ 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. $\endgroup$ Aug 12, 2019 at 20:28
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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 $X_1^2+X_2^2-1$.
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$\begingroup$ 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 $f_i$. $\endgroup$– reunsAug 12, 2017 at 10:07
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3$\begingroup$ @reuns The corollary is that maximal ideals are of the form $(X_1-a_1,\dots,X_n-a_n)$. $\endgroup$– egregAug 12, 2017 at 10:17
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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 $I = (0)$
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$\begingroup$ 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. $\endgroup$– Xander Henderson ♦Feb 21, 2019 at 17:07