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Let $R$ a commutative ring. I know that $\sqrt I=\{r\in R\mid \exists n: r^n\in I\}$, but I don't know how to compute $\sqrt I$ when $I$ is given.

For example, how can I compute $\sqrt{(x-3)}$ and $\sqrt{((x+1)^3)}$?

By the way, what does $(x,y)^2$ mean ? (it's from primary decomposition).

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Hint $\sqrt{((x-a)^n)}=(x-a)$ is an immediate consequence of the definition. (Prove it by double inclusion)

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    Shouldn't it be $\sqrt{ (x-a)}$ on the right hand side? Since we don't know whatever $x$ is. – 2017-01-28
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    @user35359 : $x-a$ is an irreducible polynomial; so by euclid's lemma, if $x-a\mid P^n$ we have $x-a\mid P$. – 2017-01-28
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    I thought we're in an arbitrary Ring $R$, not necessarily polynomial ring. – 2017-01-28
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    $\sqrt{(x-a)}=(x-a)$. To see that start from $\forall I, I\subseteq\sqrt{I}$ and $(x-a)$ is a maximal ideal – 2017-01-28
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    ok, thanks, and what is $(x,y)^2$ ? – 2017-01-28
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    $I^2=I\cdot I$ is the ideal generated by products $x\cdot y, x\in I,y\in I$ – 2017-01-28