# 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 $$A/\mathfrak{a}$$?

Could anyone help me in answering those questions please?

• 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

## 1 Answer

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.

• I do not see how I can use the first Isomorphism thm for rings? and how is this different from using the first isomorphism theorem for groups. Could you clarify the details please? Oct 2, 2022 at 0:22
• Well the one for rings is, to my memory, completely analogous to the one for groups. Here the kernel is an ideal, and the image is a ring isomorphic to the quotient of the domain ring by that ideal. Oct 2, 2022 at 0:25
• Should not I show that the image is a ring to be able to use it? Oct 2, 2022 at 0:30
• And, part of the definition of a ring is that it is an abelian group. Oct 2, 2022 at 0:33
• No, the image is a ring by the theorem. Oct 2, 2022 at 0:34