Radical of ideal: Radical of $4mathbb Z = 2mathbb Z$?

The radical of an ideal $Isubset R$ is $${rm rad}(I)={x in R mid x^n in I, n in mathbb N}.$$

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If $x in {rm rad}(4mathbb Z)$, then $x^n in 4mathbb Z$ for some $n in mathbb N$.

So, $x^n = 4m$ for some $m in mathbb Z$.

Then, $x^n = 4m = 2cdot 2m in 2mathbb Z$.

So, $2$ divides $x^n$ and hence $2$ divides $x$ (since $2$ is prime).

So, $ xin 2mathbb Z$.

Conversely, if $x in 2mathbb Z$, then $x=2m$ for some $m in mathbb Z$.

Therefore $x^2 = (2m)^2=4m^2 in 4mathbb Z$.

So, $x in {rm rad}(4mathbb Z)$.

So, ${rm rad}(4mathbb Z)=2mathbb Z$.

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In general, if $R$ is a PID and $(a)$ is an ideal, then we can write $a$ in its unique prime factorization (since every PID is a UFD), say $a=p_1^{alpha_1}cdots p_n^{alpha_n}$, and then $${rm rad}((a))=(p_1cdots p_n).$$

I am assuming that your ring in the question is $mathbb{Z}$ and you wish to calculate the radical of the ideal $4mathbb{Z}$. We will denote the radical ideal of $4mathbb{Z}$ by $sqrt{4mathbb{Z}}$. Therefore, by definition: $sqrt{4mathbb{Z}}={rin mathbb{Z}|r^nin 4mathbb{Z} text{ for some positive integer n}}$. ..(1)

Note that $2^2=4in mathbb{4Z}$ and by definition in (1), $2in sqrt{4mathbb{Z}}$ and so the ideal generated by $2$ is contained in $sqrt{4mathbb{Z}}$, and so $2mathbb{Z} subset sqrt{4mathbb{Z}}$.

To prove the reverse inclusion, take any element $rin sqrt{4mathbb{Z}}$. Suppose, we have $rin sqrt{4mathbb{Z}}implies r^n in 4mathbb{Z}$ for some positive integer $n$ and so $r^n=4m$ for some $m in mathbb{Z}$ $implies$ $r^n$ is even $implies$ $r$ is even $implies$ $r in mathbb{2Z}$ $implies$ $sqrt{4mathbb{Z}}subset 2mathbb{Z}$.And we are done.

If $R$ is a ring and $I$ is an ideal then $r(I)={xin R;|;x^nin I ;mathrm{for};mathrm{some}; n}.$ In this case, $2^2in 4mathbb Z,$ so $2in r(4mathbb Z).$