Can someone explain ideals to me?

You can think of ideals as subsets that behave similarly to zero. For example, if you will add $0$ to itself, it is still $0$, or if you multiply $0$ with any other element, you still get $0$. So ideals are like "zeros with several elements". It is a subset $I$ of a ring $R$ that is closed under addition, and also if $a\in I, b\in R$, then $ab\in I$. In particular, one single element $0$ forms an ideal in any ring $R$.

I hope this helps a bit.

An ideal is a generalization of a number that makes factoring work better if you allow square roots of some numbers.

What numbers does $n$ divide?

When studying how numbers factor, we can take a specific number, $n$, and look at all the numbers it divides, $(n) = \{ nk : k \in R \}$ where $R$ is the universe of numbers we are examining.

If you've worked with factoring problems much, you probably know these two basic rules: (1) if $n$ divides $a$ then $n$ divides $ab$ too, and (2) if $n$ divides $a$ and $b$, then $n$ divides $a+b$. The first is true, since if $a=nk$ then $ab= n(kb)$ is still a multiple of $n$. The second is true, since if $a=ni$ and $b=nj$ then $a+b=n(i+j)$, so a sum of multiples of $n$ is still a multiple of $n$.

Ideal numbers

An ideal number is a (nonempty) set of numbers that satisfies those two properties, even if we don't know what $n$ is. We can pretend there is an $n$ if we want, but later we begin to think of the ideal numbers like $(n)$ as being even better than just plain old numbers like $n$.

It turns out we can multiply ideal numbers pretty easily, we just multiply their members: if $mn$ divides $a$, then $a=mnk$ and we can write $a=m(nk)$ as a product, one from $(m)$ and one from $(n)$. Conversely, if $a=mi$ and $b=nj$ then $ab=(mi)(nj) = (mn)(ij)$ is in $(ij)$.

So we define $IJ$ to be the smallest ideal that contains all the products $ij$ for $i \in I$ and $j \in J$.

We can also find a nice analogue GCDs of numbers using ideals. If $d$ divides both $m$ and $n$, then it also divides $m+n$ and $m-n$ and $m+3n$ and $7m-4n$ and all sorts of other multiples. In other words, $d$ divides every single element of $(m) + (n) = \{ mi + nj : i,j \in R\}$. In the integers we get that $(m)+(n) = (\gcd(m,n))$, but for general rings, we just think of $I+J$ as the analogue of the GCD of $I$ and $J$.

Factoring with numbers is weird

Now it turns out there are nice rings like $R=\mathbb{Z}[\sqrt{-5}]$ where factoring numbers is very weird, $6=2\cdot 3$ cannot be factored further, but it can be factored differently, $6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$. One use of ideals is to clearly state that these are different: $(2) \neq (1\pm\sqrt{-5})$ so the way in which things are multiples of 6 is not very well described by only one of $6=2\cdot 3$ or $6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$.

Factor by any means necessary

A second use is simpler and more satisfying to me: why CAN'T we keep factoring those two until we get a common factorization? In early grade school we couldn't subtract 3 from 2, but then we invented -1 (which is like saying... but you still need to subtract 1). In later grade school, we couldn't divide 7 into 2, but then we invented $7/2$ (which is totally cheating, what is 7 divided by 2? It is 7/2 of course!). So, why not invent a new way of factoring? We just need to find the common divisor of $2$ and $1+\sqrt{-5}$. Can't do it for numbers (try it! You get $1$ as the answer, which is just useless.) But for ideal numbers it is TOOOOO easy! The GCD of the ideal numbers $(2)$ and $(1+\sqrt{-5})$ is just $$(2) + (1+\sqrt{-5}) = \{ 2i + (1+\sqrt{-5})j : i,j \in R = \mathbb{Z}[\sqrt{-5}] \}$$

We usually abbreviate that as $(2,1+\sqrt{-5})$, just leaving off the $\gcd$. Now we can refine the factorization: $$(6) = (2,1+\sqrt{-5})(2,1-\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5})$$

It turns out this is the only unrefinable way to factor the ideal number $6$ into ideal numbers, none of which is equal to the silly ideal number $(1)=R$ (which is like saying $2=1\cdot 2$ is factoring the prime number $2$).