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

Let R be a ring, and let a \in R.

  1. Prove that the set \mathsf{RAnn}_R(a) = \{ x \in R \ |\ ax = 0 \} is a right ideal and that the set \mathsf{LAnn}_R(a) = \{ x \in R \ |\ xa = 0 \} is a left ideal in R.
  2. Prove that if L \subseteq R is a left ideal, then the set \mathsf{LAnn}_R(L) = \{ x \in R \ |\ xL = 0 \} is a two-sided ideal of R.

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

  1. (x+y)A \subseteq xA + yA and A(x+y) \subseteq Ax + Ay
  2. If A \subseteq B, then xA \subseteq xB and Ax \subseteq Bx.

Proof:

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

Now to the main result.

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.

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