A proof that this set is an ideal of a commutative ring This is a homework problem which I have worked hard on, but got stuck at the last step. Any assistance would be much appreciated. The problem is from Herstein's Abstract Algebra, 3rd ed., section 4.3, p.146, #1, and reads:
If R is a commutative ring and $a \in R$, let $L(a) = \{s \in R |\ \ sa = 0\}$.
I need to prove that $L$ is an ideal of $R$. 
My solution runs along these lines:
Let $\varphi$ be the mapping from $R$ to $R$ defined by $\varphi(a) = sa, s \in R$. Then $Ker \varphi$ is the set $\{s \in R | \varphi(s) = 0 \Rightarrow sa = 0\}$. This set has been named $L(a)$ above. Thus $Ker \varphi = L$.\
Next, we will prove that $\varphi$ is a homomorphism from $R$ to itself. Let $a,b \in R$. Then $\varphi(a+b) = s(a+b) = sa + sb$, and $\varphi(a)+\varphi(b)=sa+sb$, and $\varphi(ab) = sab$, $\varphi(a)\varphi(b) = sasb = ssab$, where the rearrangement is legal because $R$ is a commutative ring, giving us that $\varphi$ is a homomorphism from $R$ to $R$. Now, since this is true, and since $L$ is the kernel of this homomorphism, then, according to Lemma 4.3.1, $L$ is an ideal of $R$. $\blacksquare$ 
Now, Lemma 4.3.1 states: If $\varphi : R \rightarrow R'$ is a homomorphism, then $Ker \varphi$ is an ideal of $R$. A proof is given in the text, but it is pretty simple.
My problem, as is probably apparent, is that I can't show that this mapping is a homomorphism over $R$'s multiplication operator, since $sab \neq ssab$. I think that I should be able to get an extra $s$ term in the expression for $\varphi(ab)$, or perhaps get rid of one of them in $\varphi(a)\varphi(b)$. What am I missing? Thank you for your time and any help you can offer.
EDIT Here is my new approach at it.
EDIT 2 Adding the important details brought up by Arturo.
In order for $L$ to satisfy the requirement of being an ideal of $R$, we need to show that $L$ is an additive subgroup of $R$ and that $rl \in L \ \forall\ r \in R\ \forall\ l \in L$. First, we will show that $L$ is an additive subgroup of $R$. It is clear that every member of $L$ is also a member of $R$, by construction, so we have $L \subset R$ to begin with.   
Next, we will show that $L$ is nonempty. This is easy to see because $0 \in R$ is such that $r0 = 0\ \forall\ r \in R$, proving that $0 \in R$ is also a member of $L$.  
Here, we will show that $L$ is closed under $R$'s addition. Let $x,y \in L$, and let $a \in R$ be the fixed $a$ in $L(a)$. The elements of $L$ have the sum $x + y$, and both are such that $xa = 0 \& ya = 0$, so $(x+y)a = xa + ya = 0 + 0 = 0$, indicating that $x + y \in L$ for any two such members of $L$, and so $L$ is closed under $R$'s addition.  
Next, we will check that every element of $L$ has an inverse. Let $x = -1*b | b \in L$. We claim that $x \in L$, which we show by first computing the product: $r*(-1) = -r$, and $r$ is such that $ra = 0$, so $(-1)ra = (-1)0 \Rightarrow -ra = 0 \Rightarrow -1 \in L$.  
Now, we have that both $-1 \& b \in L$. We claim that $rl \in L\ \forall\ r \in R\ \forall\ l \in L$. Then, since $-1 \in L \subset R$, $-1*b = -b \in L$. We will prove this claim shortly, but for now we can use this fact to show that $x = -b = b^{-1}$ by adding the two: $b + x = b + (-b) = 0$.  
These facts, together, prove that $L$ is an additive subgroup of $R$.  
Next, we will prove our claim that $rl \in L\ \forall\ r \in R\ \forall\ l \in L$. $l$ is such that $la = 0$, so the products $(rl)a = r(la) = r0 = 0$ show that $rl \in L$ as well. This proves, along with the earlier fact that $L$ is an additive subgroup of $R$, that $L$ is an ideal of $R$. $\blacksquare$
 A: Not only is your map not a homomorphism in general, you are incorrect in claiming that the kernel of the map is $L(a)$! (Also, it is very bad form to refer to the preimage of $0$ as "kernel" unless you already know that the map is a homomorphism).
In the definition of $L(a)$, $a$ is fixed. In your definition of $\phi$, $s$ is fixed. That means that the preimage of $0$ under $\phi$ is
$$\phi^{-1}(0) = \{ a\in R \mid \phi(a)=0\} = \{a\in R\mid sa=0\}.$$
So, no, this does not equal $L(a)$.
Even if you fix it by defining $\varphi_a\colon R\to R$ by $\varphi_a(s) = sa$ (which will give you that the preimage of $0$ under $\varphi_a$ is $L(a)$), you don't get a homomorphism in general: consider $R=\mathbb{Z}$, $a=2$. The map $\mathbb{Z}\to\mathbb{Z}$ given by $r\mapsto 2r$ is not a ring homomorphism.
So you want a different approach. Why not use the very definition of ideal? Clearly, $0$ is in $L(a)$. If $x,y\in L(a)$, then $xa=ya=0$, so $(x+y)a = xa+ya= 0$. And if $x\in L(a)$ and $r\in R$, then $rx\in L(a)$, because $(rx)a = r(xa) = r0 = 0$. Therefore, $L(a)$ is an ideal of $R$.

As a follow-up, let $I$ be left ideal of a not-necessarily commutative $R$. We can define the "left annihilator of $I$ in $R$" by
$$\mathcal{L}_R(I) = \{r\in R\mid ra=0\text{ for all }a\in I\}.$$
Then $\mathcal{L}_R(I)$ is a two-sided ideal of $R$. Try proving that.
A: HINT $\rm\ f(x) = a\: x\ $ is $\rm R$-linear so its kernel is an $\rm R$-module, i.e. since $\rm\ f(x-y) = f(x)-f(y)\ $ and $\rm\:f(r\:x) = r\:f(x)\:,\:$ then $\rm\:f(x) = 0 = f(y)\ \Rightarrow\ f(x-y) = f(x) - f(y) = 0\:;\ \ f(x) = 0\ \Rightarrow\ f(r\:x) = r\:f(x) = 0\:.$
A: Just for the record:
There is a natural way to realize $L(a)$ as a kernel of a map between rings.  Namely, consider the principal ideal $aR \subset R$.  It is an abelian group.  It is an ideal, and so it is closed under multipliction by $R$.  Thus there is a map $R \to End(aR)$ given
by mapping an element $r$ to the endormorphism of $aR$ induced by multiplication by $R$.  The kernel of this map consists of the elements of $R$ that annihilate every element of $a R$. It is easily seen that these are exactly the elements that annihilate $a$, and so the kernel of the map $R \to End(aR)$ equals $L(a)$.
Also:
In Bill Dubuque's answer he gives a different realization of $L(a)$ as a kernel, namely as a kernel of a map between $R$-modules.  (Because your $R$ is commutative, two-sided and one-sided ideals coincide, and so given an ideal, you can try to realize is as either a kernel of a ring homomorphism, or as the kernel of a module homomorphism.  In this particular case, the module point-of-view is probably easier, and is closer to what you were considering.  Of course, the two points of view are closely related --- e.g. in both of them the ideal (i.e. $R$-submodle) $aR$ of $R$ features prominently). 
