# Example of how a ring can have more than one maximal ideal.

I found the definition of the (Jacobson) radical interesting. It says: "Let $$R$$ be a ring. The (Jacobson) radical denoted rad$$(R)$$ is the intersection of all the maximal ideals of $$R$$.

I want to better wrap my mind around this definition because apriori, I couldn't imagine how a ring can have more than one maximal ideal. When I think of rings and ideals, I imagine this chain: $$I_1\subset I_2\subset \cdots I_n\subset R$$ where $$I_n$$ is max.

But say $$R=\mathbb{Z}$$ and $$I=(6)$$, then as $$(2)\supset(6)$$ and $$(3)\supset (6)$$ where $$(2), (3)$$ are maximal, then is this a valid example of how a ring can have multiple maximal ideals? Secondly, is a generalization of this that rad$$(\mathbb{Z})=\cap\{(p)|p \,\,prime\}$$?

• Yeah, the example with $\mathbb{Z}$ works. If you only have one maximal ideal, this is a local ring. Commented May 26, 2021 at 18:18
• Note that $\operatorname{rad}(\mathbf Z)=\{0\}$. What you you wrote is not even an ideal. of $\mathbf Z$. Commented May 26, 2021 at 18:22
• @Bernard, thanks for pointing that $rad(R)$ should be an ideal. Commented May 26, 2021 at 18:27
• Were you thinking maximal mean the same thing as maximum or something? "Having nothing above" is not the same thing as "everything is below" because the ideals are generally not linearly ordered. Commented May 26, 2021 at 18:28
• @rschwieb, in some sense yes. I was thinking of the maximal ideal to be the leading ("largest") ideal that isn't equal to the ring. Commented May 26, 2021 at 18:30

Yes, indeed $$\mathbb{Z}$$ has many maximal ideals, namely the $$(p)$$ where $$p$$ is prime. Actually, having only one maximal ideal is a very strong property, called being a local ring.
You are correct that $$rad(\mathbb{Z})=\bigcap \{(p)|p \text{ prime}\}$$, and this actually means that $$rad(\mathbb{Z})=\{0\}$$, because no non-zero integer is divisible by all primes.
• It is not true that $(p) \cap (q)=0$ when $p$ and $q$ are distinct primes. Indeed, $(2) \cap (3)=(6)$. The reason $\operatorname{rad}(\mathbb{Z})=0$ is that no nonzero integer can be divisible by every prime. I also don't understand the remark about the intersection: $\bigcap_{p \mbox{ prime}} (p)$ and $\bigcap \{(p) \mid p \mbox{ prime}\}$ seem to be slightly different notations for the same thing. Am I missing something here? Commented May 26, 2021 at 18:37