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Can you provide me some examples of an ideal $I$ of a polynomial ring $R[x]$.

I need the example for which the set defined below:

$X=\{x \in R: f(x)=0, \forall f \in I \}$

is empty. I know, that e.g. $f(x) = x^2 + 1$ could belong to such an ideal.

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    I think you are a bit confused: your set is a subset of $R$, how can a polynomial belong to it? Anyway, if you pick $I=xR[x]$, then $\forall f \in I$ you have $f(0)=0$. – 2017-02-28
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    Mentioned polynomial belongs to $I$ (which is a subset of $R[x]$), not the $R$ itself. – 2017-02-28
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    Sorry for my English! I corrected the question. – 2017-02-28
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    You can trivially take $I=R[x]$. Since $1\in I$ (or any nonzero constant polynomial), there are no common zeros of $I$. The weak nullstellensatz actually shows that this is the best you can do when $R$ is an algebraically closed field, as the set of common zeros of any proper ideal is nonempty. – 2017-02-28
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    When I take $I=R[x]$ then the set $X$ defined in the question is not empty. I need it to be empty. I corrected the initial question sorry for that. – 2017-02-28
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    @supertramp when you take $I=R[x]$, the set $X$ is empty for the reason that user Ben West explained in the first line of his previous comment. – 2017-02-28
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    oh rlly u r right! – 2017-02-28
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    any non trivial exapmle? – 2017-02-28

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Finally I found non-trivial example:

$I = $ { $f \in R[x] :f | (x^2 + 1) $ }

Then, $X$ defined in my question is empty, since $(x^2 + 1) \in I$, and $(x^2 + 1)$ has no roots in $R$. Thank you very much for discussion in comments.