I know that a monomial ideal in $k[x_1, \ldots, x_n]$, with $k$ a field, is prime if and only if is of the following type $$I = (x_{i_1}, \ldots ,x_{i_k}).$$

Is there a similar criterion to establish if a monomial ideal is primary ?


1 Answer 1


The prime monomial ideals are precisely those generated by subsets of the variables, i.e. of the form $(x_{i_1}, \ldots, x_{i_m})$, where $\{i_1, \ldots, i_m\} \subseteq \{1, \ldots, n\}$.
If $I$ is a primary ideal, then $\sqrt{I}$ is prime, so there is a subset $\{i_1, \ldots, i_m\} \subseteq \{1, \ldots, n\}$ such that $\sqrt{I} = (x_{i_1}, \ldots, x_{i_m})$. Thus some power of each of the $x_{i_j}$ is in $I$, say $x_{i_j}^{a_j} \in I$. Moreover, since $I$ is $(x_{i_1}, \ldots, x_{i_m})$-primary, the only variables involved in generators of $I$ must be the $x_{i_j}$. Thus $I = (x_{i_1}^{a_1}, \ldots, x_{i_m}^{a_m}, m_1, \ldots, m_k)$ where $m_1, \ldots, m_k$ only involve variables among $x_{i_1}, \ldots, x_{i_m}$.

Thus, a monomial ideal $I \subseteq k[x_1, \ldots, x_n]$ is primary iff every variable appearing in (a minimal generator of) $I$ has some power in $I$.

However, this does not mean that $I$ is generated by powers of variables: e.g. $(x,y)^2 = (x^2, xy, y^2) \subseteq k[x,y]$ is primary (being a power of a maximal ideal), but is not generated by powers of variables.

  • $\begingroup$ do you know some references about this topic ? $\endgroup$
    – WLOG
    Jun 15, 2014 at 19:15
  • $\begingroup$ @WLOG: See the exercises in Chapter 3 of Eisenbud, for example $\endgroup$
    – zcn
    Jun 15, 2014 at 19:27
  • $\begingroup$ @zcn : Your description was my first guess, but I thought I could be a bit greedier and I didn't really think it through to be honest. Thanks for correcting me! $\endgroup$ Jun 15, 2014 at 21:19
  • 3
    $\begingroup$ @zcn What is the proof of reciprocal implication? $\endgroup$
    – alexb
    Apr 20, 2016 at 17:06
  • 2
    $\begingroup$ This doesn't show that if $I = (x_{i_1}^{a_1}, \ldots, x_{i_m}^{a_m}, m_1, \ldots, m_k)$ as above, then $I$ is primary. How do we show this? $\endgroup$ Jun 16, 2020 at 0:07

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