Jeremy Côté


Intuition about Ideals

When studying rings in abstract algebra, one also learns about subrings. They are pretty much exactly what you would expect: subsets of a ring with the same operations defined on this subset. However, a more interesting type of ring is an ideal.

Definition: An ideal $I$ of a ring $R$ is a subring with the special property that, for any element $a \in I$ and any element $r \in R$, $ar \in I$ and $ra \in I$.

Also, note that if we have a commutative ring, then you only need to check one of the cases above. So that’s the formal definition. It’s a little clunky, but there’s a nice intuition behind it.

An ideal “absorbs” the elements that it comes into contact with. In other words, any time an element in your ring $R$ comes into contact with the an element of the ideal, it becomes part of the ideal. (I kind of want to give a zombie analogy, but I’ll let you fill in the details for now!)

On its own, this isn’t particularly interesting. So what if ideals absorb their elements?

The real magic requires a bit more theory. First, we can have a particularly kind of ideal, called a principal ideal, which is an ideal of the form $\left< a \right> ={ ar: r \in R, a \in I }$. This simply means that the element $a$ “generates” (or is a factor of) every single element in the ideal. For a quick example, if we consider the ring of integers (which is just the integers with addition and multiplication defined on them), $\left< 2 \right>$ is a principal ideal of the integers, which consists of all the even integers. No matter what integer you multiply $2$ by that isn’t in $\left< 2 \right>$, the result will be in the ideal.

We can now consider the factor ring $R/ \left< a \right> = { x + \left< a \right>:x \in R }$. I’m going to avoid talking about classes and equivalence relations here, so instead, I’ll describe the idea behind these factor rings. Essentially, the factor ring above is the set of all elements in $R$ such that any “factors” of $\left< a \right>$ are taken out. In other words, we are taking the elements $x \in R$ modulo the elements that are of the form $ay, y \in R$.

If we go back to our example with the ring of integers and $\left< 2 \right>$, we can consider the factor ring $\mathbb{Z} / \left< 2 \right>$. What are the elements which are part of this set? Well, the elements of $\left< 2 \right>$ are all of the even integers. Therefore, the elements which are left in our factor ring can’t have any factors of two in them. Furthermore, they can’t have factors of two in them as a result of division with remainder. To see this explicitly, consider $65$. This number is odd, so it isn’t divisible by two. However, $65 = 32 \cdot 2 + 1$. As such, the $32 \cdot 2 \in \left< 2 \right>$, so the ideal “absorbs” it. Operationally, that means it disappears, so we are simply left with $1$.

If you continue with this factor ring, you will see that the only numbers we can be left over with is $-1, 0$, and $1$. We can then remove $-1$ by simply adding $2$ (which is equivalent to a zero in our factor ring). What we end up with is precisely the integers modulo two.

As you can see, we these special ideals act as absorbers that take in certain elements and remove them from the ring. Why do we want to remove elements? Well, a particular type of ring that is useful to work in is a field, where all the nonzero elements are units and there are no zero-divisors. Creating a factor ring by “dividing” out certain elements generated by an ideal can give us a field. In fact, there’s a theorem called the first isomorphism theorem of rings which accomplishes precisely this function, but that’s for another time.