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Let $R$ be a prime ring. If $R$ contains a commutative nonzero left ideal, then $R$ is commutative.

my approach is that

Let $U$ be a nonzero commutative left ideal. Let $x$ in $U$ and $l$, $m$ in $R$. then $lx$ in $U$ and $mx$ in $U$. As given $U$ is commutative. $lx. mx$=$mx.lx$. Help me to show that $lm$=$ml$

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    Pick any nonzero commutative ring A and any nonzero noncommutative ring B, and consider the ring R equal to the direct product of A and B. The ideal generated by the 1 of A is commutative. – 2017-02-21
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    (There are examples which are not of this form. Any noncommutative algebra over a field with simple socle, say) – 2017-02-21
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    As all too often happens, the original question was misstated... – 2017-02-21
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    @MarianoSuárez-Álvarez Can you explain what you mean? Being a noncommutative algebra with a simple socle does not seem to be sufficient to have a nonzero "commutative left ideal": consider $\mathbb H$. I also have a different example in mind with a proper socle. – 2017-02-21

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Not true. Consider e.g. the ring of $3 \times 3$ matrices of the form $\pmatrix{a & 0 & 0\cr 0 & b & c\cr 0 & d & e\cr}$, with two-sided commutative ideal consisting of elements of the form $\pmatrix{a & 0 & 0\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$.