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

Mathematics Asked by dabofskateboarding on August 31, 2020

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 $$2$$.

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

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

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 $${rm rad}((a))=(p_1cdots p_n).$$

Answered by Juan L. on August 31, 2020

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.

Answered by Poorwelsh on August 31, 2020

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 $$2in r(4mathbb Z).$$

Answered by D. Brogan on August 31, 2020