The values of $\lambda$ for which the operations $\oplus$ and $\otimes$ are given by the following expressions.

Question

Given the following problem:

Obtain the values of $\lambda$ for which the operations $\oplus$ and $\otimes$ given by the expressions $a \oplus b = a + b - 6$ and $a \otimes b = ab + \lambda a + \lambda b + 42$ respectively, induces in $\mathbb{Z}$ the ring structure $(\mathbb{Z}, \oplus, \otimes)$. Describe in details all the steps and the justification to obtain the values of $\lambda$.

What I have tried:

• Internal Law of $(\mathbb{Z}, \otimes)$: $$ a \otimes b = ab + \lambda a + \lambda b + 42 = ab + \lambda (a + b) + 42 \in \mathbb{Z} $$ for $$ \lambda (a + b) \in \mathbb{Z} $$

• Associativity of $(\mathbb{Z}, \otimes)$: $$ a \otimes (b \otimes c) = (a \otimes b) \otimes c $$ In which I ended up with: $$ 42a + \lambda (a + \lambda c + 42) = 42c + \lambda (\lambda a + 42 + c) $$

• Distributivity: $$ a \otimes (b \oplus c) = a \otimes b \oplus a \otimes c $$ In which I got: $$ -6a - 6 \lambda = \lambda a + 36 $$ Which tells me that: $$ \lambda = -6 $$

And here I am stuck. Did I start correctly? Where do I continue?

Answer 1

The value of $\lambda$ should be an integer, because $1\otimes0=\lambda+42$ should be an integer. Note: saying $\lambda(a+b)\in\mathbb{Z}$ is wrong, because $a$ and $b$ represent variables.

It's perhaps easier to look first at distributivity (only one side because $\otimes$ is clearly commutative): \begin{align} a\otimes(b\oplus c) &=a\otimes(b+c-6)\\ &=a(b+c-6)+\lambda(a+b+c-6)+42\\ &=ab+ac-6a+\lambda(a+b+c-6)+42\\[6px] a\otimes b\oplus a\otimes c &=ab+\lambda(a+b)+42+ac+\lambda(a+c)+42-6 \end{align} After comparing both sides, we get $$ -6a-6\lambda=\lambda a+36 $$ and this should hold for every $a$. In particular, for $a=0$, we get $-6\lambda=36$, that is, $\lambda=-6$.

Now that you know the only possible value of $\lambda$, it should be easy to check all other required properties.

Answer 2

You can also see that the map $\phi:\mathbb{Z}\to\mathbb{Z}$ sending $x\mapsto x-6$ is a group isomorphism from $(\mathbb{Z},\oplus)$ to $(\mathbb{Z},+)$. Hence, you can make an Ansatz that this map is also a ring isomorphism from $(\mathbb{Z},\oplus,\otimes)$ to $(\mathbb{Z},+,\cdot)$. With this in mind, you expect $$a\otimes b-6=(a-6)\cdot (b-6)=ab-6a-6b+36\,.$$ That is, $a\otimes b=ab-6a-6b+42$. This gives $\lambda=-6$ through a sloppy method.

Remark: Since every ring structure $(\mathbb{Z},+,\circ)$ on $\mathbb{Z}$ with the additive group $(\mathbb{Z},+)$ satisfies the condition that, for some fixed $m\in\mathbb{Z}$, $a\circ b=m\,ab$ holds for all $a,b\in\mathbb{Z}$, we can justify the above Ansatz. Note that $\mathbb{Z}\circ\mathbb{Z}=\mathbb{Z}$ if and only if $m=\pm1$. Now, observe that $\mathbb{Z}\otimes \mathbb{Z}=\mathbb{Z}$ as $$a\otimes (-\lambda+1)=a-\lambda(\lambda-1)+42$$ for all $a\in\mathbb{Z}$. Hence, $\phi(\mathbb{Z}\otimes \mathbb{Z})=\phi(\mathbb{Z})=\mathbb{Z}$, and the claim follows immediately, so the sloppiness is justified.

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Alternatively, suppose instead that you are given $a\oplus b=a+b-\gamma$ and $a\otimes b=ab+\lambda a+\lambda b+\mu$, where only $\gamma$ is known. You can see that $\gamma$ is the additive identity, whence $$\gamma=a\otimes \gamma=\gamma a+\lambda a+\lambda \gamma +\mu$$ for all $a\in\mathbb{Z}$. That is, $$(\gamma+\lambda) a+(\lambda \gamma +\mu-\gamma)=0\,.$$ Since this holds for all integers $a$, $\gamma+\lambda=0$ and $\lambda \gamma +\mu-\gamma=0$, whence $\lambda=-\gamma$ and $\mu=\gamma\,(\gamma+1)$. For $\gamma=6$, it follows that $\lambda=-6$ and $\mu=42$.