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## In a ring, sets of left and right annihilators are one-sided ideals

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.22

Solution: We begin with a definition and some lemmas.

Definition: Let $R$ be a ring, $A \subseteq R$ a subset, and $x \in R$. Then $xA = \{ xa \ |\ a \in A \}$ and $Ax = \{ ax \ |\ a \in A \}$.

Lemma: Let $R$ be a ring, $A \subseteq R$ a subset, and $x,y \in R$. Then $(x+y)A \subseteq xA + yA$ and $A(x+y) \subseteq Ax + Ay$. If $A \subseteq B$, then $xA \subseteq xB$ and $Ax \subseteq Bx$.

Proof: Let $(x+y)a \in (x+y)A$. Now $(x+y)a = xa + ya \in xA + yA$. The other statement is similar. If $xa \in xA$, then $a \in B$. Thus $xa \in xB$. The other statement is similar. $\square$

Now to the main result.

(1) First, we show that $\mathsf{RAnn}_R(a)$ is a right ideal of $R$. To that end, let $x,y \in \mathsf{RAnn}_R(a)$. Note that $a0 = 0$, so that $0 \in \mathsf{RAnn}_R(a)$. Moreover, $$a(x-y) = ax - ay = 0 - 0 = 0,$$ so that $x-y \in \mathsf{RAnn}_R(a)$. Now $a(xy) = (ax)y = 0y = 0$, so that $xy \in \mathsf{RAnn}_R(a)$. Finally, if $r \in R$, then $a(xr) = axr = 0r = 0$, so that $xr \in \mathsf{RAnn}_R(a)$. Thus $\mathsf{RAnn}_R(a)$ is a right ideal of $R$.

The proof that $\mathsf{LAnn}_R(a)$ is a left ideal of $R$ is analogous.

Now let $L \subseteq R$ be a left ideal, let $x,y \in \mathsf{RAnn}_R(L)$, and let $r \in R$. Since $L0 = 0$, $\mathsf{RAnn}_R(L)$ is nonempty. Now $$L(x-y) \subseteq Lx - Ly = 0 - 0 = 0,$$ so that $L(x-y) = 0$, and $x-y \in \mathsf{RAnn}_R(a)$. Now $$L(rx) = (Lr)x \subseteq Lx = 0$$ and $$L(xr) = (Lx)r = 0r = 0,$$ so that $rx, xr \in \mathsf{RAnn}_R(L)$. Thus $\mathsf{RAnn}_R(L)$ is a two-sided ideal of $R$.