# Let $I$ be an ideal of a ring $(R,+,\cdot)$. Prove that the quotient homomorphism $\varphi : R \rightarrow R / I$ is a ring homomorphism

This is an exercise from textbook Analysis I by Amann/Escher. I would like to verify if my attempt is correct. Thank you for your help!

Let $$(R,+,\cdot)$$ be a ring with unity and $$I$$ an ideal of $$R$$. Then the following statements hold:

1. An ideal $$I$$ is proper if and only if $$1 \notin I$$.

2. A field $$K$$ has exactly two ideals: $$\{0\}$$ and $$K$$.

3. If $$\varphi : R \rightarrow R^{\prime}$$ is a ring homomorphism, then $$\operatorname{ker}(\varphi)$$ (with respect to $$+$$) is an ideal of $$R$$.

4. The intersection of a set of ideals is an ideal.

5. Let $$I$$ be an ideal of $$R$$ and let $$R / I$$ be the quotient group $$(R,+) / I .$$ Define an operation on $$R / I$$ by $$R / I \times R / I \rightarrow R / I, \quad(a+I, b+I) \mapsto a b+I$$

Show that, with this operation as multiplication, $$R / I$$ is a ring and the quotient homomorphism $$\varphi : R \rightarrow R / I$$ is a ring homomorphism.

My attempt:

1. The statement is equivalent to "An ideal $$I = R \iff 1 \in I$$"

The direction $$\Longrightarrow$$ is clear. Assume $$1 \in I$$. Then, for $$a \in R$$, we have $$a1=a \in I$$. It follows that $$R \subseteq I$$ and thus $$I =R$$. The direction $$\Longleftarrow$$ then follows.

1. Let $$I$$ be an ideal of $$K$$. If $$1 \in I$$, then for $$a \in K$$, $$a1 = a \in I$$ and thus $$I=K$$. We next consider the case where $$1 \notin K$$. Assume the contrary that there exists $$I \ni a \neq 0$$. We have $$a^{-1}a = 1 \in I$$, which is a contradiction. Thus $$I = \{0\}$$.

2. We have $$\operatorname{ker}(\varphi)$$ is a normal subgroup of $$(R,+)$$. If $$a,b \in \operatorname{ker}(\varphi)$$ then $$\varphi(a)=0'$$ and $$\varphi(b)=0'$$. It follows from $$\varphi$$ is an homomorphism that $$\varphi(ab) = \varphi(a)\varphi(b) = 0'$$. Thus $$ab \in \operatorname{ker}(\varphi)$$. So $$\operatorname{ker}(\varphi) \cdot \operatorname{ker}(\varphi) \subseteq \operatorname{ker}(\varphi)$$. As a result, $$\operatorname{ker}(\varphi)$$ is a subring of $$R$$.

Since $$R$$ is a ring with unity, $$\operatorname{ker}(\varphi) \subseteq \operatorname{ker}(\varphi) \cdot R$$ and $$\operatorname{ker}(\varphi) \subseteq R \cdot \operatorname{ker}(\varphi)$$. If $$a \in R$$ and $$b \in \operatorname{ker}(\varphi)$$ then $$\varphi(ab) = \varphi(a)\varphi(b)=\varphi(a)0'=0'$$. So $$ab \in \operatorname{ker}(\varphi)$$ and thus $$R \cdot \operatorname{ker}(\varphi) \subseteq \operatorname{ker}(\varphi)$$. As a result, $$R \cdot \operatorname{ker}(\varphi) = \operatorname{ker}(\varphi)$$. Similarly, $$\operatorname{ker}(\varphi) \cdot R = \operatorname{ker}(\varphi)$$. It follows that $$\operatorname{ker}(\varphi)$$ is an ideal of $$R$$.

1. Let $$\{I_j \mid j \in J\}$$ be a collection of ideals of $$R$$ and $$I = \bigcap_{j \in J} I_j$$. It is easy to verify that $$I$$ is a subring of $$R$$. Since $$R$$ is a ring with unity, $$I \subseteq R \cdot I$$. If $$a \in R \cdot I$$ then $$a = bc$$ for some $$b \in R$$ and $$c \in I$$. It follows that $$c \in I_j$$ for all $$j \in J$$. Thus $$a = bc \in I_j$$ for all $$j \in J$$. As a result, $$a \in I$$. So $$R \cdot I \subseteq I$$ and $$R \cdot I = I$$. Similarly, $$I\cdot R =I$$. It follows that $$I$$ is an ideal of $$R$$.

2. Since $$(R,+,\cdot)$$ is a ring, $$(R,+)$$ is an Abelian group. Thus $$I$$ is a normal subgroup of $$(R,+)$$ and $$R/I$$ is an Abelian group.

For $$a+I,b+I,c+I \in R/I$$ where $$a,b,c \in R$$, we have

\begin{aligned}((a+I) \cdot (b+I)) \cdot (c+I) &= (ab+I) \cdot (c+I)\\ &= abc + I\\ &= (a+I) \cdot ((b+I) \cdot (c+I))\end{aligned}

Then $$\cdot$$ is associative.

\begin{aligned}((a+I) + (b+I)) \cdot (c+I) &= (a+b+I) \cdot (c+I)\\ &= (a+b)c + I\\ &= (c+I) \cdot ((a+I) + (b+I))\end{aligned}

The distributive law then holds. It follows that $$(R/I,+,\cdot)$$ is a ring.

For $$a+I,b+I \in R/I$$ where $$a,b \in R$$, we have

$$\varphi (ab) =ab+I = (a+I) \cdot (b+I) =\varphi(a) \cdot \varphi(b)$$

As a result, $$\varphi$$ is a ring homomorphism.

Update: To remove any confusion arising from different definitions of the same concept, I include ones that are utilized in my textbook.

1. A pair $$(G,\odot)$$ consisting of a nonempty set $$G$$ and an operation $$\odot$$ is called a group if the following holds:

(G1) $$\odot$$ is associative.

(G2) $$\odot$$ has an identity element e.

(G3) Each $$g \in G$$ has an inverse $$g^\flat \in G$$ such that $$g \odot g^\flat = g^\flat \odot g=e$$.

1. Let $$(G,\odot)$$be a group and $$H$$ is a nonempty subset of $$G$$ that satisfies

(SG1) $$H \odot H \subseteq H$$.

(SG1) $$h^\flat \in H$$ for all $$h \in H$$.

then $$(H, \odot)$$ is itself a group and is called a subgroup of $$G$$.

1. A triple $$(R,+,\cdot)$$ consisting of a nonempty set $$R$$ and operations, addition $$+$$ and multiplication $$\cdot$$, is called a ring if

(R1) $$(R,+)$$ is an Abelian group.

(R2) $$\cdot$$ is associative.

(R3) The distributive law holds: $$(a+b) \cdot c=a \cdot c+b \cdot c \text{ and } c \cdot(a+b)=c \cdot a+c \cdot b \text{ for all } a, b, c \in R$$

1. Suppose $$R$$ is a ring and $$S$$ is a nonempty subset of $$R$$ that satisfies the following:

(SR1) $$(S,+)$$ is a subgroup of $$(R,+)$$.

(SR2) $$S \cdot S \subseteq S$$.

Then $$S$$ is itself a ring, a subring of $$R$$, and $$R$$ is called an overring of $$S$$.

1. Let $$(R,+,\cdot)$$ be a ring. A subring $$I$$ is called an ideal of $$R$$ if $$R \cdot I = I \cdot R = I$$. An ideal is proper if it is a proper subset of $$R$$.

2. $$(K,+,\cdot)$$ is called a field if the following are satisfied:

(F1) $$(K,+,\cdot)$$ is a commutative ring with unity.

(F2) $$0 \neq 1$$.

(F3) $$(K - \{0\},\cdot)$$ is an Abelian group.

In 5, the most important part is to check that addition and multiplication are well defined on $$R / I$$ - that if $$a + I = b + I$$ and $$c + I = d + I$$ then $$(a + I) + (b + I) = (c + I) + (d + I)$$ and $$(a + I) \cdot (b + I) = (c + I) \cdot (d + I)$$.
For addition: if $$x \in (a + I) + (b + I)$$ then $$x = a + u + b + v$$ for some $$u,v \in I$$. If $$a + I = c + I$$ and $$b + I = d + I$$ then $$a + p = c + q$$ for some $$p, q \in I$$, and so $$a = c + \alpha$$ for some $$\alpha \in I$$, analogously $$b = d + \beta$$. Then $$x = c + (\alpha + u) + d + (\beta + v)$$. As $$I$$ is closed under addition, $$\alpha + u \in I$$ and $$\beta + v \in I$$, so $$x \in (c + I) + (d + I)$$. Can you make similar prove to addition? (this is where you need $$I$$ to be an ideal and not just a subring)
Other parts are correct, but can be shortened. I think it will be useful to prove that you don't need to prove ideal be subring: any subset $$I$$ that is additive subgroup s.t. $$IR\subseteq I$$ and $$RI \subseteq I$$ is automatically a subring.
You also don't actually need unity in $$R$$ in 3 and 4. For example, in $$4$$, to show $$RI \subseteq I$$ we can simply use $$R\cdot I = R\cdot (\cap I_j)$$ (definition), $$R\cdot (\cap I_j) \subseteq \cap (R \cdot I_j)$$ (general property: $$f(X, \cap Y_i) \subseteq \cap f(X, Y_i)$$ and $$\cap (R \cdot I_j) \subseteq \cap I_j$$ (all $$I_j$$ are ideals).