If $I$ and $J$ are ideals in $R$, and $I$ is a subset of $J$, is $I$ also an ideal in $J$?

Question

Pretty much what the title suggests.

If $I \subset J$ are both ideals in a commutative ring $R$, is it true that $I$ is an ideal in $J$?

My reasoning for this is that clearly for all $a,b\in I$, $a - b \in I$ and since any element that is in $J$ is also in $R$, $I$ should also be closed under multiplication by $J$.

Something tells me this isn't right but I can't see why. Any help would be appreciated!

Answer 1

Yes.

More is true: If $S$ is a subring of $R$, and $I\subset S$ is an $R$-ideal, then it is also an $S$-ideal.

This is the case for exactly the reasons you say.

Answer 2

Think of ideal as additive groups. Consider $I \subset J \subset R$. We already know $I$ has additive group structure in $R$. Let $x \in I$ and $j \in J \subset R$. Then, $xj \in I \subset J$. Lastly $x,y \in I \Rightarrow x,y \in J$, moreover $x+y \in I \subset J$ and these are all the properties to conclude $I$ is an ideal of $J$.

Answer 3

I recommend that you don't use the phrase "ideal in"; use the phrase "ideal of". Why? Because it is meaningless to say that some set $X$ "is an ideal" without defining a context, i.e., what ring is it an ideal of. The set of even integers $$I=\{\ldots,-4,-2,0,2,4,\ldots\}$$ is an ideal of the ring $\mathbb{Z}$, but it is not an ideal of the ring $\mathbb{Z}[x]$. It has no ideal-nature that is independent of what ring we're talking about.

Now that that's out of the way, note that whenever you say $$X\;\;\textsf{ is an ideal of }\;\; Y$$ then $Y$ must be a ring (of course, in addition to $X$ being a subset of the ring $Y$ satisfying the relevant properties).

If your definition of "ring" requires that any ring must have a multiplicative identity, then in many situations, an ideal of a ring is not itself a ring. As an example, $\mathbb{Z}$ is a ring, and $I$ (the set of even integers) is an ideal of $\mathbb{Z}$, but $I$ does not have a multiplicative identity. Thus, it makes no sense to ask whether some subset of $I$ is, or is not, an ideal of $I$ because $I$ is not a ring.

However, if your definition of "ring" does not require that any ring must have a multiplicative identity, then it is true for any ring $R$ and any ideal $J\subseteq R$ that $J$ is itself a ring and that any ideal $I$ of the ring $R$ that happens to be contained in $J$ will itself be an ideal of $J$-considered-as-a-ring.