Ideals

Think of an ideal as certain types of set of integers which is closed under addition and multiplication. The concept of an ideal will pay big dividends for us in the next chapter.

Definition 1.4.1   Let $ I\subset \mathbb{Z}$ be a subset closed under addition and subtraction, that is to say, for any $ r,s\in I$, $ r+s\in I$ and $ r-s\in I$. Such a subset is called an ideal of $ \mathbb{Z}$.

Note you can multiply any element of $ I$ by any integer and the product will still be in $ I$. You can add two elements (but they both must belong to $ I$) and the sum will still be in $ I$.

Thanks to the following result, the answer to the first question asked above is ``yes''.

Lemma 1.4.2   Let $ I\subset \mathbb{Z}$ be a subset closed under addition and subtraction Then either

The proof below is worthwhile trying to understand well since it contains a basic idea which occurs in other parts of mathematics (in fact, we shall see the idea again in the next chapter).

proof: If $ I$ is not empty then it must contain 0. Suppose that $ I$ is non-zero. Let $ r\in I$ be non-zero. Since $ -r\in I$, and either $ r>0$ or $ -r>0$, $ I$ must contain a positive element. By the well-ordering principle, $ I$ contains a least element $ a>0$.

Claim: $ I=a\mathbb{Z}$.

If $ b\in I$ then the division algorithm (theorem 1.2.7) says that there is an $ r$ such that $ 0\leq r<a$ with $ b=qa+r$. Since $ I$ is closed under addition and subtraction, $ r=b-qa$ belongs to $ I$. But $ a$ is the smallest non-zero element of $ I$, so $ r$ must be zero. This implies $ b\in a\mathbb{Z}$. Since $ b$ was choosen arbitrarily, this imples $ I\subset a\mathbb{Z}$. The reverse inclusion $ a\mathbb{Z}\subset I$ follows from the assumption that $ I$ is closed under addition and subtraction. This proves the claim and the lemma. $ \Box$


David Joyner 2007-09-03