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 $R$. $\blacksquare$