# Proof step in Steps in Commutative Algebra

I am trying to solve this exercise in Sharp's Steps in Commutative Algebra (I did not copy the question itself, just the first part of the exercise to prevent solutions of it): For me we should have $a-b\notin I^t$ above the red line.

I need help.

• $a - b$ is an element of the ring R. $I^t$ is subset of $R$. then what is the meaning of $a - b \neq I ^ t$ ? Jul 20 '13 at 7:12
• @GA316 I'm sorry, my mistake Jul 20 '13 at 7:16
• @GA316 Now, it's ok. Jul 20 '13 at 7:18
• The book is OK. There is no greatest $t$ such that $a-b$ is not in $I^t$; for all sufficiently large $t$, $a-b$ is not in $I^t$. Jul 20 '13 at 7:37
– Did
Jul 20 '13 at 8:27

I think your mistake lies in negating the sentence of the form $\exists x : P(x)$ is true". The expanded form of the sentence in the book is

There exists a greatest $t_0 \in \Bbb{N}$ such that $(a-b) \in I^{t_0}$.

What is the negation of this sentence? Well first we need to understand what it means! It means that for any $t > t_0$, we have $(a-b) \notin I^t$. Thus the negation of the sentence in the book is:

There is no greatest $t_0 \in \Bbb{N}$ such that $(a-b) \in I^{t_0}$.

In other words, here is the situation that is going to happen. Start with $a-b$ which is an element of $R = I^0$. Note by definition $I^0$ is equal to $R$. The must be some $t_1 > 0$ such that $(a-b) \in I^{t_0}$ otherwise this $0$ would be the greatest element in $\Bbb{N}$ such that $(a-b) \in I^0$ and $a-b \notin I^t$ for all $t > 0$.

Having chosen $t_1$, we can again choose another $t_2$ such that $a-b \in I^{t_2}$ again by repeating the argument in the paragraph before. Continuing this process ad infinitum shows we have a sequence of natural numbers

$$t_0,t_1,t_2,\ldots$$

such that $a-b \in I^{t_k}$ for all $k \geq 0$. In summary:

The sentence "There is no greatest $t_0 \in \Bbb{N}_0$ such that $a-b \in I^{t_0}$" implies the sentence "There is a sequence $t_0,t_1,\ldots$ such that $a-b \in I^{t_k}$ for all $k \geq 0$.

Assume there is no greatest $t$ such that $a-b\in I^t$, then $a-b\in I^{t_k}$ for some increasing sequence $\{t_k:k\in\mathbb{N}_0\}$. In this case $$a-b\in\bigcap\limits_{k=1}^\infty I^{t_k}\subset\bigcap\limits_{n=1}^\infty I^n$$ so $\bigcap_{n=1}^\infty I^n\neq 0$. Contradiction, hence there is the greatest $t\in\mathbb{N}_0$ such that $a-b\in I^t$.