My algebra teacher asked whether the ideal $(X_1, X_2, \dots, X_n) $ can be generated by fewer polynomials over the field $K[X_1, X_2, \dots, X_n]$.

My intuition tells me that it can't, so I tried to suppose the opposite. If it could, then there would be $P_1, P_2, \dots, P_{n-1} \in K[X_1, X_2, \dots, X_n]$ such that $(P_1, P_2, \dots, P_{n-1}) = (X_1, X_2, \dots, X_n)$ (if fewer than n-1 polynomials suffice, I could just pick some more out of ${X_1, X_2, \dots, X_n}$).

It follows that there are some polynomials $Q_{ij} \in K[X_1, X_2, \dots, X_n],\space i \in \{1, \dots n\},\space j \in \{1, \dots n-1\}$ such that:

$P_1 \times Q_{1,1} + P_2 \times Q_{1,2} + \dots + P_{n-1} \times Q_{1,n-1} = X_1$


$P_1 \times Q_{n,1} + P_2 \times Q_{n,2} + \dots + P_{n-1} \times Q_{n,n-1} = X_n$

Here I kinda got stuck so I would appreciate any help. :)

  • 5
    $\begingroup$ Denote the ideal $\left(X_1,X_2,...,X_n\right)$ by $M$, and the ring $K\left[X_1,X_2,...,X_n\right]$ by $A$. Then, $M/M^2$ is an $A/M$-vector space (this is a general fact, and proven completely straightforwardly) of dimension $n$. (The ring $A/M$ is a field and isomorphic to $K$.) Now what would happen if you had $n-1$ generators for $M$? $\endgroup$ – darij grinberg Mar 27 '12 at 22:28
  • $\begingroup$ @darijgrinberg Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 9 '13 at 21:18
  • $\begingroup$ Related: math.stackexchange.com/questions/422182/… $\endgroup$ – Watson Nov 25 '16 at 21:04
  • $\begingroup$ Somehow related: math.stackexchange.com/questions/95004/… $\endgroup$ – Watson Dec 22 '16 at 17:26

Your intuition is right: The ideal $\left(X_1,X_2,...,X_n\right)$ of $K\left[X_1,X_2,...,X_n\right]$ cannot be generated by less than $n$ elements.

Proof: Let $M$ be the ideal $\left(X_1,X_2,...,X_n\right)$ of $K\left[X_1,X_2,...,X_n\right]$. We need to show that this ideal $M$ cannot be generated by less than $n$ elements.

Assume the contrary. That is, the ideal $M$ can be generated by less than $n$ elements. In other words, the $R$-module $M$ can be generated by less than $n$ elements.

But the ring $R/M$ is isomorphic to the field $K$ (why?), hence a field. Thus, $R/M$-modules are $R/M$-vector spaces.

Generally, if $I$ and $J$ are two ideals of a commutative ring $R$, then $J/IJ$ is an $R/I$-module (with the action defined in an obvious way: $\overline r \cdot \overline j = \overline{rj}$, where $\overline r$ means the residue class of $r\in R$ modulo $I$, where $\overline j$ means the residue cass of $j\in J$ modulo $IJ$, and where $\overline{rj}$ means the residue class of $rj\in J$ modulo $IJ$). Applied to $R = K\left[X_1,X_2,...,X_n\right]$, $I = M$ and $J = M$, this yields that $M/M^2$ is an $R/M$-module. Since the $R$-module $M$ can be generated by less than $n$ elements, the $R/M$-module $M/M^2$ can be generated by less than $n$ elements (for instance, the projections of the less than $n$ generators of $M$ onto $M/M^2$). Since $R/M$-modules are $R/M$-vector spaces, this rewrites as follows: The $R/M$-vector space $M/M^2$ has dimension $< n$. Hence, its $n$ elements $\overline{X_1}$, $\overline{X_2}$, ..., $\overline{X_n}$ are linearly dependent (over $R/M$). In other words, there exist elements $a_1$, $a_2$, ..., $a_n$ of $R$ such that $a_1X_1 + a_2X_2 + ... + a_nX_n \in M^2$ but not all of $a_1$, $a_2$, ..., $a_n$ lie in $M$ (why?). Consider such elements.

Since $a_1X_1 + a_2X_2 + ... + a_nX_n \in M^2$, the coefficient of $a_1X_1 + a_2X_2 + ... + a_nX_n$ before $X_1$ equals $0$ (because every polynomial in $M^2$ has its coefficient before $X_1$ equal $0$). But the coefficient of $a_1X_1 + a_2X_2 + ... + a_nX_n$ before $X_1$ is clearly $a_1\left(0\right)$ (since the only term in the sum $a_1X_1 + a_2X_2 + ... + a_nX_n$ which can contribute to the coefficient before $X_1$ is the first term). Thus, $a_1\left(0\right) = 0$. In other words, $a_1 \in M$ (since $M$ is the set of all $P\in K\left[X_1,X_2,...,X_n\right]$ satisfying $P\left(0\right)=0$). Similarly, $a_i \in M$ for all $i\in\left\lbrace 1,2,...,n\right\rbrace$. Thus, all of $a_1$, $a_2$, ..., $a_n$ lie in $M$. This contradicts the fact that not all $a_1$, $a_2$, ..., $a_n$ lie in $M$. This contradiction finishes the proof.


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