What is the kernel of a homomorphism? I've done a lot of problems before but I am trying to get a really basic   definition of kernel so that I may apply to any possible given question that I may be presented with.  
Would  I be correct in saying that the kernel (of a  homomorphism) is basically what I can multiply any given function by to get the identity?
 A: To the OP: the other answers have given you the definition of the kernel of a group homomorphism only, probably because you tagged this question with group theory.
If you are familiar with ring theory (if not, you will be soon), we can have a homomorphism from a ring $R$ to a ring $S$.  But a ring has both an additive identity, usually denoted $0$, and a multiplicative identity, usually denoted $1$.  In this special case, the kernel of the homomorphism is defined as the stuff in $R$ that is mapped to $0$.
Also, if you have ever taken a linear algebra course, and know about linear transformations between vector spaces, the kernel of a linear transformation is the stuff in the domain that is mapped to the $0$ vector in the co-domain.
A: If you're avoiding symbols, you could call it the preimage of the identity (of the target group), and think of it as "everything that gets sent to the identity (of the target group)".
A: Yes, sort of. The kernel of a group homomorphism $\phi:G\to H$ is defined as
$$
\ker\phi=\{g\in G:\phi(g)=e_H\}
$$
That is, $g\in\ker\phi$ if and only if $\phi(g)=e_H$ where $e_H$ is the identity of $H$.
It's somewhat misleading to refer to $\phi(g)$ as "multiplying $\phi$ by $g$". Rather, we use the language "applying $\phi$ to $g$" to emphasize that $\phi$ is a function between groups not an element of one of the groups in question.
Example. Note that $\Bbb Z$ and $\Bbb Z^2$ are groups under addition. Moreover, the identities are $e_{\Bbb Z}=0$ and $e_{\Bbb Z^2}=(0,0)$.
Let $\phi:\Bbb Z^2\to\Bbb Z$ be the group homomorphism defined by $\phi(a,b)=a+b$. Then $(a,b)\in \ker\phi$ if and only if $\phi(a,b)=0$. That is, $(a,b)\in\ker\phi$ if and only if $a+b=0$. Hence $(a,b)\in\ker\phi$ if and only if $b=-a$.
This proves that $\ker\phi=\{(a,-a):a\in\Bbb Z\}$.
As noted in the comments, kernels arise in lots of other contexts. If you're interested, see the "mathematics" section of the wikipedia entry for kernel.
If you're feeling extra ambitious, you could learn category theory and see how the kernel of a group homomorphism is a special case of an equalizer.
