Radical of an ideal equals the intersection of all prime ideals containing it.

Question

I am trying to prove that for a ring $R$ and an ideal $I\leq R$ we have $$ \sqrt{I}=\bigcap_{I\leq\mathfrak p}\mathfrak p, $$ the intersection of all prime ideals containing $I$.

The definition of the radical of an ideal I am working with is $$ \sqrt{I}=\{x\in R\mid\exists m\in\mathbb{N} \text{ with }x^m\in I\}. $$ I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?

Answer 1

I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$.

Following through with definitions, you can prove $\sqrt{I}\subset N$.
Proving the reverse containment is not as simple. This shall be done by proving the contraposition: $x\notin \sqrt{I}\Rightarrow x\notin N$.
Given $x\notin \sqrt{I}$, consider the collection $\Omega$ of all ideals $J \supset I$ such that for all $n\in \Bbb N$, $x^n \notin J$. Partially order $\Omega$ by inclusion, and show by Zorn's lemma that $\Omega$ has a maximal element $\frak{p}$.
Since $\mathfrak{p}\supset I$, if you can show that $\frak{p}$ is prime, then you can claim $x\notin N$ (since $x\notin \frak{p}$).
Take $a,b\notin \frak{p}$, and using maximality of $\mathfrak{p}$, show that $\mathfrak{p} + (ab) \notin \Omega$; this will imply $ab\notin \mathfrak{p}$ and thus $\mathfrak{p}$ is prime (this proves the contraposition of $ab\in\mathfrak{p}\Rightarrow a\in\mathfrak{p}\vee b\in\mathfrak{p}$).

Answer 2

The one sentence answer is, for any $x \notin \sqrt{I}$, consider the ring homomorphism

$$R \to (R/\sqrt{I}) \to (R/\sqrt{I})_x$$ pick any maximal ideal in $(R/\sqrt{I})_x$ it pulls back to a prime ideal of $R$ containing $\sqrt{I}$ and not containing $x$, which is what we wanted to find.

This answer is in some sense the same as the other given answer by Kobe but hides the Zorn's lemma argument in the fact that any nonzero ring has a maximal ideal:

Here is the argument in excessive detail:

To avoid having to write $\sqrt{I}$ all the time, let $J=\sqrt{I}$ so that $J$ is radical ideal (if $x^n \in J$ then $x \in J$). If $\mathfrak{p}$ is a prime ideal containing $I$ then it contains $J$ (if $x \in J$ then $x^n \in I$ so $x^n \in \mathfrak{p}$ so $x \in \mathfrak{p}$). So we want to show $$J \supseteq \bigcap_{J \subseteq \mathfrak{p}} \mathfrak{p}$$ where the intersection is over prime ideals.

As in Kobe's proof we show this by showing if $x \notin J$ then $x \notin \bigcap \mathfrak{p}$, which means we have to find a prime ideal $\mathfrak{p}$ containing $J$ but not containing $x$.

So assume $x \notin J$. The key insight is that $J$ being radical means $(R/J)_x$ is nonzero ring: if we had $0=1$ in this localization, then by definition of equality in localization, $\overline{x}^n\cdot 1=0$ in $R/J$ so $x^n \in J$, and since $J$ is radical, $x \in J$ contrary to assumption.

Now consider the composite of the ring canonical homomorphism $$ R \to R/J \to (R/J)_x$$ Since $(R/J)_x$ is nonzero ring, it has a maximal ideal $\mathfrak{m}'$ (using Zorn's lemma), and consider its preimage $\mathfrak{p}$ in $R$...it need not be a maximal ideal (although it will be if $R$ is finite type over a field...that's some form of a bunch of results that go by Zariski's lemma) but it will a prime ideal since pre-images of prime ideals are primes and maximal ideals are prime. And by the corresponding between prime ideals in a ring and it quotients/localizations, $\mathfrak{p}$ is prime ideal of $R$ containing $J$ and not containing $x$ which is what we wanted to find.

Answer 3

Let $x\in\sqrt{I}$. Then $x^n\in I$. Pick an arbitrary $\mathfrak{p}$ containing $I$. Then $x^n\in I\subset\mathfrak{p}\implies x\in\mathfrak{p}$.

For the other inclusion, suppose for the sake of contradiction $x\in\bigcap\limits_{\mathfrak{p}\in specR\\ I\subset\mathfrak{p}}\mathfrak{p}$ and $x\notin\sqrt{I}$. We may suppose $1\notin I$, else $1\in I\implies I=R\implies \sqrt{I}=\sqrt{R}=R$ and the inclusion is trivial. So we have that $S=\{1,x,x^2,x^3,\dots\}\subset R\setminus I$ is multiplicatively closed and therefore $R\setminus S$ is prime and contains $I$. Then $x\in R\setminus S$ and we have our contradiction.