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Let $R$ be a ring and $S$ be an ideal of $R$.

Let $Ann_R(S)_r=\{a\in R | Sa=0\}$ and $Ann_R(S)_l=\{a\in R | aS=0\}$ denote the right annihilator and the left annihilator of $S$ in $R$, respectively.

If $R$ is a semiprime ring (i.e Given an ideal $S$, a semiprime ring is one for which $S^n=0$ implies $S=0$ for any positive $n$), then $Ann_R(S)_r=Ann_R(S)_l$.

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$A=Ann_R(S)_rS$ is an ideal.

$A^2=\{0\}$.

If $R$ is semiprime, this means $A=\{0\}$.

This means $Ann_R(S)_r\subseteq Ann_R(S)_l$.

By symmetry, the reverse containment holds.

QED.

One can also compare this to the obvious element-wise version that holds for noncommutative reduced rings: if $xy=0$ then $yx=0$. Proof: $(yx)^2=0$, so $yx=0$.

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    Is $I$ an arbitrary ideal? – 2017-01-18
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    @Ninja Sorry, you're right, I was thinking $S$. There is no need for "let". I am just making an assertion. – 2017-01-18
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    "Let" was needed to make a edit containing $6$ characters. – 2017-01-18
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    @Ninja well, FFR don't make edits that necessitate corrections later just to dodge character limits. It's much better just to leave it as a suggestion in the comments in that case. – 2017-01-18