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

Question

The definition says that $x^n$ is in the ideal but I don't see what positive integer $4$ can be raised to in order to get $2$.

Answer 1

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

Answer 2

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

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

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

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

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

So, $ x\in 2\mathbb Z$.

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

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

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

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

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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 $${\rm rad}((a))=(p_1\cdots p_n).$$

Answer 3

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

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

To prove the reverse inclusion, take any element $r\in \sqrt{4\mathbb{Z}}$. Suppose, we have $r\in \sqrt{4\mathbb{Z}}\implies r^n \in 4\mathbb{Z}$ 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 $\implies$ $r \in \mathbb{2Z}$ $\implies$ $\sqrt{4\mathbb{Z}}\subset 2\mathbb{Z}$.And we are done.