Nilradical of a graded ring Let $S_\bullet$ be a graded and commutative ring with unity. Recall that the nilradical, the ideal of nilpotent elements of $S_\bullet$, can also be defined as 
$$
\mathcal{N}(S_\bullet)=\bigcap_{p\lhd_pS_\bullet} p,
$$
where by $p\lhd_p S_\bullet$ I mean that $p$ is a prime ideal of $S_\bullet$. Can the nilradical be found by intersecting only graded prime ideals? For this question, an ideal is graded if it is generated by homogeneous elements. 
Alright, so not every prime ideal is graded, therefore we have
$$
\bigcap_{p\lhd^H_p S_\bullet}p \supseteq \bigcap_{p\lhd_p S_\bullet }p,
$$
where $p\lhd_p^H S_\bullet$ means that $p$ is graded prime ideal. On the other hand, denote by $p^H$ the ideal generated by homogeneous elements of $p$. It is clear that $p^H$ is graded and $p^H\subseteq p$. Therefore, we have 
$$
\bigcap_{p\lhd^H_p S_\bullet}p \supseteq \bigcap_{p\lhd_p S_\bullet }p\supseteq \bigcap_{p\lhd_p S_\bullet }p^H.
$$
But on the other hand, not every graded prime ideal is obtained by homogenising a prime ideal, so if we throw in the rest of the graded prime ideals, we get a smaller intersection.
$$
\bigcap_{p\lhd^H_p S_\bullet}p \supseteq \bigcap_{p\lhd_p S_\bullet }p\supseteq \bigcap_{p\lhd_p S_\bullet }p^H\supseteq \bigcap_{p\lhd^H_p S_\bullet}p.
$$
Therefore, 
$$
\mathcal{N}(S_\bullet)=\bigcap_{p\lhd_p^H S_\bullet}p. 
$$
Is that right? The result seems very counterintuitive to me and I could not find any references for it.
 A: Your proof seems correct to me. I would in fact point out that once you replace every prime ideal with its homogenization, you actually get the intersection of all homogeneous prime ideals, since the homogeneous ideals were already there and are equal to their homogenization. In other words, the last step of adding the prime ideals that are already homogeneous is not needed. 
Now let me try to convince you that this result is not so counterintuitive, by showing that the nilpotency of an element is governed by the nilpotency behavior of its homogeneous components. 
Suppose that we have an element $a = a_s + a_{s+1}+ \cdots a_t$ of $S$ which is nilpotent, where $a_i$ is its $i$th homogeneous component. This means that $a^n=0$ for some positive integer $n$. Expanding the power, we can see that $a_s^n$ is precisely the component of degree $ns$ of $a^n$. In particular $a_s^n=0$ and so $a_s$ must be nilpotent. Since both $a$ and $a_s$ are nilpotent so is $a-a_s$, and in a similar way we can show that $a_{s+1}$ is nilpotent. By induction we see that every homogeneous component $a_i$ of $a$ must be nilpotent as our wise teacher user26857 had asserted. 
