0

Let $\mathfrak{a},\ \mathfrak{p}\subset R$ be two ideals, $\mathfrak{p}$ prime, with $\mathfrak{a}\not\subset\mathfrak{p}.$ Show that $\mathfrak{a}^n\not\subset\mathfrak{p}$.

I tried this by induction on $n$.

So $n=1$ is just the hypothesis, supose that $\mathfrak{a}^{n-1}\not\subset\mathfrak{p}$. We can consider all the $\alpha^2\in\mathfrak{a}^n$ such that $\alpha\in\mathfrak{a}^{n-1}$. If all the $\alpha^2$ were on $\mathfrak{p}$ we would have that $\alpha^2=\alpha\ ยท \alpha\ \in\mathfrak{p}$ but $\mathfrak{p}$ is prime so $\alpha\in \mathfrak{p},\ \forall\alpha\in\mathfrak{a}^{n-1},$ which is a contradiction.

Is my prove correct? Is there any other prove? Am i just trying to prove somethig trivial?

Somerandommathematician
  • 200

2 Answers2

2

Let's show the contrapositive: if $\mathfrak{a}^n \subseteq \mathfrak{p}$, let $a\in \mathfrak{a}$ and note that $a\cdot a^{n-1}=a^n\in \mathfrak{a}^n \subseteq \mathfrak{p}$. By primeness we have either $a\in \mathfrak{p}$ and/or $a^{n-1}\in \mathfrak{p}$. If the former is true we are done, if the latter holds we can, again conclude, by primeness, that either $a\in \mathfrak{p}$ and/or $a^{n-2}\in \mathfrak{p}$. Keep on applying this argument til you eventually have $a\in \mathfrak{p}$, showing that $\mathfrak{a} \subseteq \mathfrak{p}$.

Jonathan Hole
  • 3,564
0

You can modify your argument slightly to get a correct proof: we will exhibit an element of $\mathfrak a^n \setminus p$ for all $n$. We have $\mathfrak a \not\subseteq \mathfrak p$ by assumption; pick $\alpha \in \mathfrak a \setminus \mathfrak p$ once and for all. We modify the induction hypothesis to be that $\alpha^{n} \in \mathfrak a^n \setminus \mathfrak p$ (which of course implies that $\mathfrak a^n \not\subseteq \mathfrak p$, i.e., what we ultimately want). The case $n=1$ is trivial. Now in the induction step, clearly $\alpha^n \in \mathfrak a^n$, but $\alpha^n \notin \mathfrak p$, since otherwise either $\alpha$ or $\alpha^{n-1}$ is in $\mathfrak p$, both of which we know to be false by induction.

user555203
  • 3,367