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.


1 Answer 1


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).


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