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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 $R$. $\blacksquare$

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Try to prove directly that $L(a)$ is an ideal: show it is an additive subgroup and that $xy\in L(a)$ whenever $x\in R$ and $y\in L(a)$. You are making this more complicated that it is! – Mariano Suárez-Alvarez Nov 20 '11 at 6:16
Also, if you define $\varphi(s)=sa$, then $\{s \in R | \varphi(s) = 0 \Rightarrow sa = 0\}$ is all of $R$, because the condition $\varphi(s) = 0 \Rightarrow sa = 0$ is always true (everything implies itself, and the only difference between the left and the right side of the implication is the unfolding of $\varphi$). – Henning Makholm Nov 20 '11 at 7:01
Thanks to you as well @Henning, my general intent with adding that was to only write out the unfolding explicitly. I clearly should not have done so, given the logical interpretation you have brought up! – karmic_mishap Nov 20 '11 at 19:03
Thanks to both @MarianoSuárez-Alvarez and Arturo Magidin for your suggestions to approach the problem more directly. I did so at first and must have messed something up badly, because I abandoned that route and tried this tack instead. Now that I have tried it again, it seems much simpler. I will add my new approach to my question now. – karmic_mishap Nov 20 '11 at 19:26
@karmic_mishap: You are missing just two tiny details in your proof: in order for $L$ to be an additive subgroup you need to show that it is closed under addition, but also that is it nonempty and closed under inverses. The latter will come "for free" if you prove the second condition of being an ideal, since if $a\in L$ then $(-1)a = -a$ will be in $L$; so you can get away with only checking sums, but you cannot call it an "additive subgroup" until you are done with the entire proof. But more importantly, you need to check it is nonempty, which you never did. – Arturo Magidin Nov 20 '11 at 22:30
up vote 4 down vote accepted

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.

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Thank you for your help. I was a bit worried about my calling it a kernel as well, I should be able to remember that in the future. It seems that everything I wrote about this seemingly simple map was indeed quite muddled. I appreciate the continued patience in pointing that out. – karmic_mishap Nov 20 '11 at 19:01

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 $\rm\:f(x) = 0 = f(y)\ \Rightarrow\ f(x-y) = f(x) - f(y) = 0\:;\ \ f(x) = 0\ \Rightarrow\ f(r\:x) = r\:f(x) = 0\:.$

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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 $L(a)$.


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).

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I like your reply, but now I need to go learn what an endomorphism is. Thanks for the new math, really! According to your reply, it must be related as the left annihilator mentioned elsewhere. I will have to look into both of them. – karmic_mishap Nov 22 '11 at 5:33
Just a non-invertible automorphism! Wish the textbook included this stuff. Any other good ones you would recommend? – karmic_mishap Nov 22 '11 at 5:35
@karmic: Dear karmic_mishap, There are lots of algebra texts available. I've taught from Dummit and Foote before, and it seemed good (at least, its exercises are good). Lots of people like Artin. There is also the tome by Lang. You may want to get hold of a few such common texts and dip into each of them. Regards, – Matt E Nov 22 '11 at 5:48

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