Show that the ideal $((x_1-\alpha_1)^{t_1},...,(x_n-\alpha_n)^{t_n})$ is primary.

Question

Show that for all choices of $t_1,...,t_n \in \mathbb{N}$, the ideal $$Q = ((x_1-\alpha_1)^{t_1},...,(x_n-\alpha_n)^{t_n})$$ of $R=K[x_1,\dots,x_n]$ is primary.

What I've done: we have that $Q \subseteq M$, where $M = (x_1-\alpha_1,...,x_n-\alpha_n)$ is a maximal ideal. I'll show that $R/Q$ is non zero and every zero divisor of $R/Q$ is nilpotent.

Let $a = f + Q, b = g + Q \in R/Q$ such that $ab = 0 $ but $b \neq 0$ (i.e, $g \notin Q$). Since $fg \in Q \subseteq M$, and $M$ is maximal then prime, we have $f \in M$ or $g \in M$. I was able to show that if $f \in M$ then there is a power $k$ of $f$ such that $f^k \in Q$, that is, $a$ is nilpotent.

My quest is: How can I show, from $g \notin Q$, that $g \notin M$?

(This is the exercise 4.38 from Sharp)

Answer 1

Let $a,b\in R$, and suppose $ab\in Q$.

We want to show $a\in Q$ or $b^m\in Q$, for some positive integer $m$.

Let $k=\max\{t_1,...,t_n\}$.

Then $M^{nk} \subseteq Q \subseteq M$.

In particular, if $b\in M$, then $b^{nk}\in Q$, and we're done.

Suppose $b\not\in M$.

It remains to show $a\in Q$.

From $M^{nk} \subseteq Q \subseteq M$, we get $\text{rad}(Q)=M$.

Hence $M$ is the only prime ideal containing $Q$.

Since $b\not\in M$, it follows that the ideal $(Q,b)$ is not contained in any maximal ideal, hence \begin{align*} &(Q,b)=(1)\\[4pt] \implies\;&a(Q,b) = (a)\\[4pt] \implies\;&a\in a(Q,b)\\[4pt] \implies\;&a\in a(Q+(b))\\[4pt] \implies\;&a\in aQ+(ab)\\[4pt] \implies\;&a\in Q\\[4pt] \end{align*} as required.

Answer 2

To show that the ideal is primary, you need to show that $f\in M$ using that $g \not \in Q$. You cannot prove that $g \not\in M$ as you can see with some simple examples (take $Q=(x^2)$ in $K[x]$ and $f=g=x$).

So in the case that $g\in M$, assume for the sake of contradiction that $f\not \in M$. Let $h$ s.t. $fh-1\in M$ (inverse modulo $M$). We know that $fh-1$ is nilpotent modulo $Q$ so $(fh-1)^k\in Q$, which proves that there exists $h'$ s.t. $fh'-1\in Q$. So we have $fg \in Q$, and $fgh'-g\in Q$ so $g\in Q$, which is a contradiction.