0

Suppose $R(\oplus,\odot)$ is a ring without unity. If $\mathbb{Z}(+,.)$ is the usual ring of integers, and if we define the ring $\mathbb{Z}\times R$ with the ring operations $++$ and $\times$ as $$(m,a)++(n,b)=(m+n,a\oplus b)$$ $$(m,a)\times(n,b)=(m.n,m\odot b\oplus n\odot a\oplus a\odot b)$$ where $m,n\in\mathbb{Z}$ and $a,b\in R$.

Then, is the ring $\mathbb{Z}\times R$ with the operations described a ring with unity?

Though it is easy to see that it is a ring, but I am having a hard time in finding the unity element. The non-commutativity of the $\odot$ operation makes it look difficult. Any hints?

Jyrki Lahtonen
  • 140,891
vidyarthi
  • 7,315
  • 1
    What do $m\odot b$ mean when $m\in \Bbb Z$ and $b\in R$? Is it just $b\oplus b\oplus\cdots\oplus b$ with $m$ copies of $b$? – Arthur Apr 27 '24 at 11:18
  • @Arthur It is not upright mentioned as such. Does that assumption make it any simpler? – vidyarthi Apr 27 '24 at 11:21
  • 2
    Try $(1,0)$ :-) This construction is straight out of textbook(s), and thus is also likely covered in earlier threads on this site. At least Jacobson describes it (and I don't much care for the rest of them, but probably many/most of them have it, too). – Jyrki Lahtonen Apr 27 '24 at 11:24
  • @JyrkiLahtonen I also lean towards that, but should we beforehand declare $1\odot a=a$? – vidyarthi Apr 27 '24 at 11:28
  • 1
    If $R$ had a multiplicative neutral element $1$, and we equated an integer $m$ with the sum of $m$ copies of $1$ (ditto with $ma$), then, by the distributive laws $$(m+a)(n+b)=mn+am+na+ab,$$ which is a big hint as to the origin of this recipe! – Jyrki Lahtonen Apr 27 '24 at 11:30
  • 1
    The only interpretation of $m\odot a$ that makes sense here is to think of it as the sum of $m$ copies of $a$, using the existing addition in $R$. Once you get to modules (if not sooner), you learn this operation satisfies all the natural "rules". – Jyrki Lahtonen Apr 27 '24 at 11:33

0 Answers0