0
$\begingroup$

I read somewhere that if $I,J$ are ideals of a ring $R$ with $I \subset J$, then $R/I$ "contains" $R/J$. Please give a formal explanation of what this means, and briefly why it is true? Thanks

  • 2
    Well, that doesn't seem quite a smart thing to say. There is a morphism $R/I \to R/J$ that's onto (can you find it ?) but not necessarily an embedding $R/J \to R/I$ (there is sometimes an additive group embedding, for instance when $R$ is a principal domain), so to say that it contains it seems odd. For instance, there is no (ring-)embedding from $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$ (indeed, any morphism has to send $1$ to $1$, and thus it would send $0$ to the same thing as $1+1$, so $2$, but in the latter ring, $2\neq 0$) – 2017-02-08
  • 0
    pepa.dvorak : the way you've defined it, one has $J\subset I$, not the other way around. – 2017-02-08
  • 0
    @Max I think what you're talking about in your first comment is probably the best explanation for the poster: could you apply it in a solution please? (Well, maybe not the first sentence, but the second sentence.) – 2017-02-08

1 Answers 1

2

So as I was saying in the comments, usually when someone says that $X$ is contained in $Y$ when $X$ and $Y$ are two similar structures, and when "$X\subset Y$" is obviously false, it means that there is an embedding, i.e. an injective morphism $X\to Y$ (the type of morphism depending on the kind of structure we're dealing with).

I said that it wasn't quite smart to say so in this case since there is (generally speaking) no embedding $R/J \to R/I$, it is thus misleading to say that "$R/J$ is contained in $R/I$" (for a counterexample, see my first comment)

However, there is an onto morphism $R/I \to R/J$, making $R/J$ a homomorphic image of $R/I$. This morphism is quite simple to find : as $I \subset J$, we have for $x,y\in R$, $x\simeq y (mod I) \implies x\simeq y (mod J)$ (where I let $\simeq$ denote the congruence modulo a certain ideal). Thus, letting $[x]^I$ denote the class of $x$ mod $I$ (similarly for $J$), the morphism $R/I\to R/J$, $[x]^I \to [x]^J$ is well defined (checking that it is a morphism is routine, I can do it if you want me to) and obviously onto.

Usually, as homomorphic images aren't (in general) isomorphic to any subring, saying "contains" seems odd.

I'll put a second "however" on top of that : if you haven't given us all the information, then there is a way in which $R/I$ might "contain" $R/J$: as I've mentioned, in particular cases (for instance if $R$ is a principal domain, I haven't thought about any other examples), there may be a group-embedding from $R/J\to R/I$. In such a case, it is "smart" to say that $R/I$ as an additive group "contains" $R/J$ as an additive group. In the example I had given with $R= \mathbb{Z}, I = 4\mathbb{Z}, J= 2\mathbb{Z}$, then although there is no ring embedding $R/J\to R/I$ (and indeed, no ring morphism), there is a group embedding (just send $1$ to $2$). If you want me to, I can generalize this to $R$ being any principal domain.

EDIT : I also forgot to mention why the onto morphism $R/I \to R/J$ "felt natural " : as has been mentioned by other comments, since $I\subset J$, it is readily seen that $J/I $ is an ideal in $R/I$, and so taking the quotient of $R/I$ by $J/I$, we get $R/I / (J/I) \simeq R/J$. For those who are reading this answer and who know a bit about Universal Algebra, this is a special case of a theorem about congruences, saying that if $\Theta_0$ is a finer congruence on the algebra $A$ than $\Theta_1$, then $(A/\Theta_0 )/ (\Theta_1/\Theta_0) \simeq A/\Theta_1$