# A smart way of proving that $\langle A/\mathfrak{a}, +, . \rangle$ is a ring.

Here is the proposition I have:

Let $$f: A \to B$$ be a ring homomorphism, then $$\operatorname{ker}f$$ is an ideal of $$A,$$ and $$\operatorname{Im}f$$ is a subring of $$B.$$ And we have $$A/\operatorname{ker}f \cong \operatorname{Im}f.$$

And here are the definitions of the operations $$+$$ and $$.$$ in $$\langle A/\mathfrak{a}, +, . \rangle :$$

If $$x \in A,$$ then the cosets of $$A/\mathfrak{a}$$ are of the form $$x + \mathfrak{a}.$$ Now, define, $$(x + \mathfrak{a}) + (y + \mathfrak{a}) = (x + y) + \mathfrak{a}$$ And $$(x + \mathfrak{a})(y + \mathfrak{a}) = xy + \mathfrak{a}.$$ Now, I want to prove that $$\langle A/\mathfrak{a}, + \rangle$$ is an abelian group in a smart way,

1- Can I say that it is an abelian group by the first Isomorphism theorem for groups? also,

2- Is the ordinary way of proving it is just by showing closure and associativity of $$+$$ and the existence of identity and inverse for every element in $$A/\mathfrak{a}$$?

• If $G$ is an abelian group, and $H$ is a subgroup of $G$, then $G/H = \{g+H:g \in G\}$ is also an abelian group. Oct 2, 2022 at 0:02
• @azif00 Can you please mention the theorem that says this? Nov 7, 2022 at 9:58

How about using the first isomorphism theorem for rings? And the fact that $$(A/\mathfrak a,+)$$ is the abelian group of the quotient ring $$(A/\mathfrak a,+,\cdot)$$?
The answer to $$2)$$ is clearly yes. That's I guess what I should say is any way of doing is going to be equivalent to that.