Questions about homomorphisms?

Question

I'm running into algebra using homomorphisms for A LOT of things, but I don't think I have a full understanding of what they are. I have read on this site a good explanation that was really great, but I feel like I still lack something.

From what I read, ring homomorphisms show how alike two structures are and the kernels of the homomorphisms are said to say how much alike the two structures are. Whether they be rings or fields or groups.

The relationship that has been causing me havoc is the one that relates the kernel of a homomorphism to the ideal of that ring. What relation does a subset like that have with homomorphisms. Im not asking for a proof,there are plenty out there. I'm just asking for a justification and understanding of homomorphisms and what they are. I want to understand them on a more intuitive level, but it is giving me trouble.

Answer 1

I may be interpreting your question wrong, but you seem to be asking for an intuitive explanation for why the kernels of ring homomorphisms behave the way they do.

Intuitively since everything in the kernel maps to 0, the elements of the kernel should behave like 0. For example $0+0=0$, so if $i,j \in I=\ker\varphi$ for some ring homomorphism $\varphi$, then $i+j$ should also "behave like 0", i.e. $i+j$ should be in $I$. Therefore $I$ should be additively closed. Since $0$ always behaves like 0 and $-0=0$, we would also expect $0\in I$ and $-i\in I$. Therefore using this intuitive argument, we expect $I$ to be an additive subgroup.

Now ideals have one more special property, closure under left and/or right scaling. Where does this come from? It comes from 0s property that $0\cdot a=a\cdot 0=0$, so we would expect that $a\cdot i$ and $i\cdot a$ would be in $I$ as well when $I$ is the kernel of a ring homomorphism.

All this intuition is what gets formalized in the proofs you've read.

Finally I suppose I should just mention that the reason ideals have no other properties (other than what you can derive from these) is that for any two-sided ideal $I$, we can construct a ring homomorphism from say $R$ to $R/I$ which has $I$ as its kernel. In other words, every kernel has these properties, and every set with these properties can be a kernel, so we don't expect any more properties.

I hope I interpreted your question right, and that this is useful.

Answer 2

A homomorphism is a function/map from one algebraic object to another of the same kind that preserves the algebraic structure on the objects. For rings, a homomorphism preserves the addition and multiplication (and possibly the identity). The kernel of such a ring homomorphism $f: A \rightarrow B$ is an ideal of $A$ and the smaller it is, the more alike the two rings are. In the extreme case, if the kernel is $(0)$, you can think of $A$ as a subring of $B$ because the map is injective. The other extreme is if the kernel is all of $A$, meaning $f$ is the $0$-map. It may be that $A$ and $B$ are so "different" that there are no non-zero homomorphisms $f: A \rightarrow B$. Also, given an ideal $I$ of $A$, you can define a homomorphism to a ring so that the kernel is $I$, namely the natural quotient map $A \rightarrow A/ I$.

An ideal $I$ of a ring $A$ is by definition a subset of a ring that is itself also a ring and closed under multiplication by elements of the larger ring $A$. Given a homomorphism $f: A \rightarrow B$ between rings, $ker(f)=\{a \in A|f(a)=0\}$. The kernel of a ring homomorphism is an ideal because a homomorphism preserves the ring structure (you can add two elements in the kernel, "subtract", multiply and multiply by elements in $A$). Compare this with the kernel of a linear map between vector spaces.

Answer 3

If you have two sets with the same type of algebraic structure on them (for example, two rings), a homomorphism is a map between the sets which preserves the structure. So, it's the simplest way to look at relationships between algebraic objects.

If you have a ring homomorphism $f: R \to S$, we define the kernel of $f$ as the part of $R$ that gets collapsed to $0$ in $S$. You can also define subsets of a ring called ideals.

Given a ring, there are usually lots of homomorphisms from that ring to other rings. And, there are usually lots of ideals in the ring.

And it turns out that kernels and ideals are basically the same thing. The kernel of a homomorphism is always an ideal. (You should check this if it's not clear.) And, going the other way, any ideal is the kernel of some homomorphism. (If $J$ is an ideal in $R$, there's a ring $R/J$ and a homomorphism $R \to R/J$.) They're just two different ways of looking at the same concept.

Maps from a ring can collapse parts of the ring, and ideals are the parts that can get collapsed.