I have edited my question according to your suggestions and comments.

Suppose now that we have a ring $R$ and a subring $S$ of $R,$ so we can define the Jacobson radicals, for example as the ideal consisting of those elements in $R$ (of $S$) that annihilate all simple right $R$-modules ($S$-modules). this radicals can be characterized in terms of maximal ideals in case that $R$ or $S$ are unital rings, and in this case $J(S)\subseteq S$.

So suppose that $R$ has unity and $S=I$ is an ideal. Then $J(I)$ Jacobson radical of $S$ is the intersection of all maximal ideals of $R$ containing $I.$ So $J(S)\supseteq S.$

This is what make me confused because with one definition $J(S)\subseteq S$ and with the other definition $J(S)\supseteq S.$ Recall also that the Jacobson radical is hereditary, that is, for any ideal $I$ of $R,$ one has that $J(I)=J(R)\cap I,$ so $J(I)\subseteq I.$

I will mantain the previous question:

Let $R$ be a ring and $I$ an ideal of $R.$ I have found that the $J(I)$ Jacobson radical of $I$ is the intersection of all maximal ideals of $R$ containing $I.$ In this case do we have $J(0)=J(R)$

This definition makes me confuse, why $J(I)$ is not the intersection of all maximal ideals of $I$ ?

Any reference will help.

  • $\begingroup$ For $I$ is in general a non-unitary ring and possibly has no maximal ideals? $\endgroup$
    – user26857
    Oct 23, 2015 at 15:01
  • $\begingroup$ What do you mean "maximal ideals of $I$?" Definitions are for usefulness, and it is not clear what you alternative would be useful or why you find the current definition not useful. $\endgroup$ Oct 23, 2015 at 15:03

2 Answers 2


I think you're blurring two notions of "radical": the radical of an ideal and the radical of a ring.

  1. A radical of an ideal $I$ is often defined by taking the intersection of special ideals containing $I$.

  2. The radical of a ring is often defined from a radical of an ideal by saying "The radical of $R$ is defined to be the radical of the zero ideal in $R$."

So: in your first paragraph you are looking at a radical of an ideal, and in your second statement, you are thinking of a radical of a $I$ considered as a ring.

Another illustration for commutative rings is $\sqrt{I}:=\cap\{P\mid P\text{ prime and } I\subseteq P\}$. This is called the radical of $I$, and the radical of the ring $R$ is $\sqrt{0}$, the radical (nilradical, more commonly) of the ring.

You can do something analogous with modules and submodules.

Generally speaking, though, the notation $J(-)$ is not standardly employed for radicals of ideals, but it is usually used radicals of rings/modules. If you use $J(-)$ to denote a radical of an ideal with the definition you used, then $J(R)=R$ in every case, because the set of maximal ideals containing $R$ is empty, and the empty intersection of ideals is usually interpreted as the whole ring. More standardly, $J(R)$ denotes the intersection of maximal ideals containing $\{0\}$, and this can be equal to all of $R$ in rings without identity. It's not $R$ if $R$ has identity, though.


To expand on the comments so far, your definition does not guarantee the existence of $J(I)$. For example, take $R=k[x_1,x_2,\ldots]$ to be a polynomial ring in infinitely many variables. Then, $I=(x_1,x_2,\ldots)$ has no maximal ideals.

The reason $J(I)$ is defined the way that it is is because it is the preimage of $J(R/I)$ under the correspondence theorem, making it very natural.


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