Basic questions in Algebra Could someone confirm or check these out? Some of these are really simple, but I can't find a definite answer to these.
Let $R$ be a ring, $D$ an integral domain, $F$ a field, $G$ a group, and $N$ a normal subgroup of$G$


*

*Do rings have to be closed under multiplication? My book only says multiplication needs to be associative.

*Is this a proof that fields are integral domains? Let $x,y \in F - \{ 0 \}$. Assume $xy = 0 \iff xy + 1 = 1 \iff xy + xx^{-1} =1 \iff x(y + x^{-1}) = 1 \iff x^{-1} = y + x^{-1} \iff y = 0$, a contradiction, so no divisors of $0$, meaning $F$ is a $D$

*Can someone briefly explain to me why the (first) Isomoprhism  theorems want to base on the Kernel so much? Without going the details, my intepretation of the theorems tells me we are just trying to cut up the group $G$ into ways we can understand it better thruogh the quotient, but I don't get why the general groundwork can't be used for arbitrary normal groups, why the kernel? My book does all these proofs, but I am not getting why the kernel is so special, why can't the isomoprhism theorems be based on some other normal groups. This is basically a case of "why" even though I know the "how".
 A: 1) Yes, rings have to be closed under multiplication.
2) That is a proof, but you can go simpler. Assume $xy = 0$. If $x = 0$, we're done. Otherwise $x \neq 0$ and so $x^{-1}$ exists. Then $y = x^{-1}xy = (x^{-1})(0) = 0$.
3) The First Isomorphism Theorem essentially states that given a homomorphism $\psi \colon G \to H$, then $G/\operatorname{ker}\psi \equiv \operatorname{im}\psi \leq H$. Basically, we use the kernel because it contains information about how $G$ is embedded into $H$. If you want the result with any other normal subgroup $N$, you'll have to find a homomorphism that has $N$ as the kernel. And there is, but it is trivial. Simply take $H = G/N$ and define $\psi(g) = gN$. 
A: I'll give my take on these


*

*Closure is an axiom in general for any structure and may be omitted simply because it is so obvious it must be closed. If we have a structure where $a,b\in R$ but $ab\notin R$ then is the definition of multiplication meaningful? Not at all because multiplication inherently is a function from $R\times R\to R$ and as such $ab$ must also be in $R$ for it to be a genuine multiplication morphism.

*Yes, your proof and the other fellows demonstrates it, in many cases we even define that a field is an integral domain with inverse elements for all non-zero elements.

*The reason is quite simply that the kernel of a homomorphism tells us things about the general behavior of the homomorphism. This is because of the axioms that a homomorphism must satesfy in order to be a homomorphism. For example let's assume that $\varphi(a)=\varphi(b)$, this implies that $\varphi(a)-\varphi(b)=0 = \varphi(a-b)$ which means that $a-b\in\ker \varphi$, this also work with multiplication notion for a group. This however is the definition of equivalence in a quotient group and quotient ring and as such it gives us that information.

A: The Quotient Set G/H is made into a group by the multiplication
aHbH = abH
and
aHbH = abb^-1HbH = ab(b^-1Hb)H = abH
only if 
(b^-1Hb)  is a subgroup of H, that is,  only if H is normal., 
The kernel of any groupp hom is a normal subgroup of the domain; and any normal subgroup is the kernel of some group hom.
