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The book I have first goes over group theory. Once it gets to rings and starts discussing subrings along with cosets and factor rings it leaves out some details for brevity and to not repeat what has already been said in the group theory portion. But in this instance, it left something out or perhaps it can be easily derived.

It defines an ideal as a subring $H$ of a ring $R$ such that $\forall a \in R$ it is the case that $aH, Ha \subset H$, and it then states this is the analogue of the definition of normal subgroup in the context of group theory and implies a factor ring $R/H$ is created by the existence of some natural homomorphism.

Here is where I'm confused. It makes no mentions that the criterion for being a normal subgroup also has to be satisfied in order for this factor ring to be well-defined. Am I correct that the criteron for being a normal subgroup has to also be satsified in order to a (1) be an ideal and (2) in order that a factor ring can be formed?

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    $\begingroup$ The additive group of a ring is abelian, so all subgroups are normal. $\endgroup$ – Daniel Fischer Oct 21 '14 at 14:58
  • $\begingroup$ To be totally clear, when you're talking about an object with two operations, it'd be bets to mention what binary operation you're thinking of. From context we can guess you probably mean the addition operation, but it's also nearly as plausible that another student would ask the same question but be thinking of the multiplication operation (with some misconceptions.) $\endgroup$ – rschwieb Oct 21 '14 at 15:26
  • $\begingroup$ I'm just learning this stuff. (Groups) For $G/H$ to be well defined, $H$ has to be a normal subgroup (left and right cosets equal). (Rings) extending the idea to rings, for $R/N$ to be well-defined, $N$ has to be a "normal subring," but we don't call it that, we call it an Ideal. We can't call $N$ a normal subgroup of $R$ because we're in the domain of rings, not groups, and the definitions are different (although "normal subring" seems more suggestive to me). It's analogous to forming a well defined object $A/B$, just using different structures. At least that's how I understand it so far. $\endgroup$ – Pixel May 17 at 5:50
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Yes, as an ideal is closed under addition and subtraction, it is a subgroup. Since a ring is moreover an abelian group under addition, every subgroup is normal.

EDIT: To explicitly answer your questions:

1) Since a subring is also a subgroup and addition is abelian, something must first be a normal subgroup in order to be an ideal.

2) Yes, well-definedness under addition does technically require being a normal subgroup.

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    $\begingroup$ I think this misses the point. The question seems about the multiplicative structure. $\endgroup$ – quid Oct 21 '14 at 15:08
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    $\begingroup$ If that is the case, then it should be noted that since rings are not groups under multiplication, there are no subgroups in the first place. $\endgroup$ – Jacob Bond Oct 21 '14 at 15:11
  • $\begingroup$ I think this answer is correct. It is more a question about what was glossed over in the book and not made explicit. I think the first comment on the question also clarifies this. I forgot that the additive group of a ring is abelian and all subgroups of abelian groups are normal (at least by the definitions in my book) so normalcy is there as soon as you start talking about subrings. $\endgroup$ – user172428 Oct 21 '14 at 15:14
  • $\begingroup$ @TheOldHag I might also mention that normal subgroups are not always ideals. For example, $\mathbb{Z}$ in $\mathbb{R}$ is a normal additive subgroup, but not an ideal. In general, subrings of fields are not ideals. $\endgroup$ – Jacob Bond Oct 21 '14 at 15:29
  • $\begingroup$ Okay. Sorry for the remark, then. I still feel the point might be missed and added my own answer. $\endgroup$ – quid Oct 21 '14 at 15:58
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Every ideal is the kernel of a ring homomorphism. Since every ring homomorphism is an additive homomorphism. every ideal is the kernel of an additive homomorphism, and so a normal additive subgroup.

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    $\begingroup$ The question seems about the multiplicative structure. $\endgroup$ – quid Oct 21 '14 at 15:08
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Recall the situation for a group $G$ and a subgroup $H$.

We know if $H$ is a normal subgroup then one gets a factor group $G/H$, if it is not normal then not.

But let us recall more precisely what is the issue there. If $H$ is not normal one can still define the set $G/H$ (and also $H\backslash G$) but does not anymore have an induced composition law on it. So, $G/H$ is not a group, yet not "for the usual reason" that some of the propoerties of the composition law is not fulfilled but since there is no reasonable law to begin with.

Note if you have classes $gH$ and $g'H$ and you want to define $(gH) \cdot (g'H) = gg'H$ then you need $ghg'h' \in gg'H$ for all $h,h' \in H$, which leads to the condition $hg' \in g'H$ and further to $g'^{-1}hg' \in H$, that is exactly the property for $H$ being a normal subgroup.

Now, let us turn to rings. You have a ring $R$ and a subring $H$ and you wonder when can one define a ring-structure on $R/H$ (where $R/H$ is defined as classes $\mod H$ with respect to addition). For the additive structure there is never a problem as $(H,+)$ is a subgroup of $(R,+)$ and by commutativity it is normal, and everything is fine.

However, what about multiplication. How to define $(a+H) \cdot (b+H)$. Well, 'of course' as $ab+H$. But for this to be well-defined you need that $(a+h)(b+h') \in ab+H$ for all $h,h' \in H$ which is the case if (and in general only if) $ah' \in H$ and $hb \in H$, that is you need $aH \subset H$ and $Hb \subset H$. And, note that you also need that additional condition for commutative rings! (Of course then one of the two suffices.)

So, if you want a multiplication on $R/H$ then you can only consider $H$ such that $aH \subset H$ and $Ha \subset H$ for all $a \in R$ (and also $HH \subset H$ but this is given by being a subring and/or as a special case of the just mentioned property), just like when you want a multiplication on $G/H$ then you can only consider $H$ such that $g^{-1}Hg \subset H$ for all $g \in G$.

This analogy is what is pointed out there.

To answer you questions more specifically, the properties you need for $R/H$ to have a natural ringstructure are:

  1. $(H,+)$ is a [normal] subgroup of $(R,+)$.

  2. $HH \subset H$.

  3. $aH \subset H$ and $Ha \subset H$ for all $a \in R$.

The first two are summarized as saying $H$ is a subring (but in fact you could drop 2 altogether as it is a special case of 3).

However, the reason that normal subgroups are mentioned there at that point is not 1. but that 3. is a condition analogous to "normal" in the group context.

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To make a quotient of $(R,+)$ by a subgroup $(H,+)$ and get the group $(R/H,+)$ you will need $(H,+)$ to be normal in $(R,+)$. Luckily since $(R,+)$ is an abelian group, all subgroups are normal.

In terms of the multiplicative structure, $(H,\cdot)$ need not be normal in $(R,\cdot)$. (You may see immediately that the lack of multiplicative inverses forces you to reconsider the definition of "normal".) But since $R/I$ will never be group with the multiplication operation given by $R$, it is not important for $R/I$ to be a multiplicative group. It is a monoid, though.

Here's an example of what can happen with the multiplicative monoid of the ring. Consider the ring $R=F\langle x,y \rangle$ is noncommuting variables and $F$ is a field. Look at the quotient $S=R/(xy,x^2)$. then $xS=xF$, but $yx\in Sx$. This demonstrates that $xS\neq Sx$. (I sidestepped the fact that many elements of $S$ won't have multiplicative inverses.)

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Theorem: Let $A$ be an ideal of ring $R$. Then $A$ is a normal subgroup of $R$.

Proof Idea: We will show that ideals are themselves abelian subgroups, we will then show all abelian subgroups are normal.Therefore by hypothetical syllogism ideals are normal subgroups.

Proof:Since an ideal is a subring, $A$ is itself $a$ ring under the operations of $R$. Since, by definition, rings are themselves abelian groups under addition, the ideal $A$ is itself an abelian group under addition. Now since $A$ is both an abelian group under addition and a subset of $R$, it is an abelian subgroup of $R$.

Now we appeal to the theorem: If $G$ is an Abelian group, then every subgroup of $G$ is normal. (Recall the definition of a normal subgroup as one wherein the left coset = the right coset). We can use the normal subgroup test, (which can also be considered to be a definition of a normal subgroup), to prove this theorem. This test states the following:

A subgroup $H$ of $G$ is normal in $G$ if and only if $$\forall x\in G, xHx^{-1} \subseteq H $$ Note that $xHx^{-1}= \{xhx^{-1} :h\in H\}$

We apply the normal subgroup test by letting $x\in G,h\in H$. Then we see: $$xhx^{-1}=hxx^{-1}=he=h \in H $$ since $h,x$ were arbitrary, we see that this is true for all $h,x.$

Therefore, since we previously showed the ideal A was an abelian subroup of R, and we now showed abelian subgroups are normal, we have shown ideals are normal subroups. $\square$

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