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And also, another thing, I'm curious about. They say that an Ideal is the analogue of the Normal subgroup in group theory, but that confuses me.

Let a Group be G. Let a subgroup be H. H is normal in G if and only if, $gH = Hg$, for all $g$ in $G$.

An Ideal:

Let $<R,+,*>$ Be a ring with set $R$ and the usual operations of addition and subtraction. Let $I$ be a subset of $R$. We say that $I$ is an ideal if it forms a subgroup $<I,+>$ of $<R,+>$ under addition. And also, for $a,b$ that is in $I$:

$a*b$ is in $I$ and $b*a$ is in $I$.

So a subgroup $H$ is normal if and only if for all elements $aH = Ha$, but a SET is an ideal if the following conditions hold. I don't understand the analogue of the two. An ideal requires that multiplication be defined and in our set I. Normal subgroups require something different. What am I missing?

user121615
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  • See also the discussion here. – Dietrich Burde May 05 '15 at 18:24
  • The second condition is for a subring. For a (two-sided) ideal, the condition is: $$$$For all $a\in I$, and all $, b\in\color{red}R, …$. – Bernard May 05 '15 at 18:24
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    The integers are a subring of the rationals, or reals. – André Nicolas May 05 '15 at 18:30
  • I know but an Ideal is a set, not a ring. While a Normal subgroup is a group. One has an operation defined on it, the other doesn't. How can that be analogous? Is the subring that it forms the analogy? – user121615 May 05 '15 at 18:32
  • What I don't understand is why they just don't say that I is a subring? Why define it as a set? They must have a reason.Would a good way to look at an ideal be that an Ideal is a set that can form a subring? – user121615 May 05 '15 at 18:36
  • An ideal is an additive subgroup, and in fact the same as an $R$- submodule of $R$. The analogy is maybe most clear in the respective properties being equivalent to the existence of a group resp. ring structure on the quotient (which makes the projection a homomorphism). The problem with subrings is that they are usually assumed to contain $1$. – Thomas Poguntke May 05 '15 at 18:38
  • Haha, I don't understand what R submodule of R is. That is definitely beyond my ken. An additive subgroup? Surely that's not the only thing it is. It also has another requirement that the set is also closed under multiplication. Why can't one look at it as a subring? – user121615 May 05 '15 at 18:41
  • Well, an $R$-submodule is precisely that: an additive subgroup which is closed under multiplication, but even by elements of $R$. You can see above an example of a subring which is not an ideal ($\mathbb Z \subseteq \mathbb Q$). The aforementioned property is not fulfilled. In fact, the smallest ideal containing $\mathbb Z$ is $\mathbb Q$ itself. (which is a rather trivial statement since $\mathbb Q$ is a field) But this holds for any "subring" of a ring, in the sense that it contains 1. – Thomas Poguntke May 05 '15 at 19:16
  • Okay, so in your particular case, you have got a subring which is indeed not an ideal. But is every ideal a subring? I guess that's what I'm trying to ask – user121615 May 05 '15 at 19:31

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