Question

Question 15:

1. Let R be the set of integers equipped with the usual addition and multiplication given by ab = 0for all$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$. Show that R is a commutative ring.
2. Show that$\mathbf{M}\mathbf{=}\left\{0,\pm 2,\pm 4,\pm 6,\cdots \right\}$ is a maximal ideal in R that is not prime. Explain why this result does not contradict Corollary 6.16.

Short answer

Answer:

It is proved that R is a commutative ring

Step-by-step solution

Commutative ring

A commutative ringwould be a ring that satisfy the following axiom:

ab = ba for every$a,b\in R$ [Commutative multiplication]

 Step 2: Show that R is a commutative ring

This implies that$\left(R,+\right)$ would be an abelian group because Rand z have the same addition.

Multiplication is associative as follows:

$$\begin{gathered} \left(ab\right)c=0c \\ =0 \\ =a0 \\ =a\left(bc\right) \end{gathered}$$

Multiplication is commutative as follows:

$ab=0=ba$

Multiplication spreads over the sum as follows:

$$\begin{gathered} a\left(b+c\right)=0 \\ =0+0 \\ =ab+ac \end{gathered}$$

As a result, R is a commutative ring.

Hence, it is proved that R is a commutative