Can the ideal $(X_1, X_2, \dots, X_n) $ be generated by fewer polynomials over the field $K[X_1, X_2, \dots, X_n]$?

Question

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:

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

$\dots$

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

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

Answer 1

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.