Understanding the proof that the ideal $(X_1, X_2, \dots, X_n) $ cannot be generated by fewer than n elements

Question

I am trying to understand the marked answer in this post: Can the ideal $(X_1, X_2, \dots, X_n) $ be generated by fewer polynomials over the field $K[X_1, X_2, \dots, X_n]$?

In particular, I would like to ask for clarification of the final paragraph: 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.

I am not sure what the replier meant by : every polynomial in $M^2$ has its coefficient before $X_1$ equal $0$ and thus $a_1$ has to be 0.

Edit: I would like to request the moderators do not marked this post as a duplicate and refer that link because I am precisely asking for clarification about the proof in that post.

Answer 1

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 $0$)... Thus, $a_1\left(0\right) = 0$.

As you should have suspected, the phrase "the coefficients of $a_1X_1 + a_2X_2 + ... + a_nX_n$ before $X_1$" can be misleading.

That phrase means the coefficient of $X_1$ in the polynomial $B(X_1,X_2,\cdots,X_n)$, where $B(X_1,X_2,\cdots,X_n)= a_1X_1 + a_2X_2 + ... + a_nX_n$. It does NOT mean $a_1$.

For example, let $n=2$, $a_1=3+5X_2$, $a_2=X_2$. Then $$B(X_1,X_2)=(3+5X_2)X_1+X_2X_2=3X_1+5X_2X_1+X_2^2.$$ That phrase means $3$, the coefficient of $X_1$ in the term "$3X_1$", instead of $a_1=3+5X_2$. Note that $3=a_1(0)$.