If I have two ring structures $(R,+,\cdot)$ and $(T, +', \cdot')$. Let $S = R \times T$ and define addition $+^{*}$ and multiplication $\cdot^{*}$ on $S$ as:

Addition: $(r_{1}, t_{1}) +^{*} (r_{2}, t_{2}) = (r_{1} + r_{2}, t_{1} +' t_{2})$

Multiplication: $(r_{1}, t_{1}) \cdot^{*} (r_{2}, t_{2}) = (r_{1} \cdot r_{2}, t_{1} \cdot' t_{2})$

If I let the additive identity be $0_{S} = (0_{R}, 0_{T})$ and the the additive inverse be $-s = (-r, -t)$ where $-r$ and $-t$ are the additive inverses for $R,T$ respectively. Also let the multiplicative identity $1_{S} = (1_{R}, 1_{T})$ where $1_{R}$ and $1_{T}$ are the multiplicative identities for $R,T$ respectively.

Then does $(S, +^{*}, \cdot^{*})$ form a ring?

^This was a question in an Abstract Algebra textbook that I'm currently reading. I have gone through the axioms for Rings and I can't see any reason why it would not be true. Can anyone validate whether that is correct or not?


  • 2
    $\begingroup$ Yes, it's a ring. You should also verify associativity and distributivity, though. $\endgroup$
    – egreg
    Mar 27, 2018 at 8:53
  • $\begingroup$ I did, the way they posed the question in the textbook indicated it may not be correct and I've been rattling my brain to check and re-check and re-check to see if I'd missed something simple. Thanks for your comment - has put my mind at ease! $\endgroup$
    – user150203
    Mar 27, 2018 at 9:03

1 Answer 1


It is correct. This procedure works very generally; for example, it will work for any algebraic structure that can be expressed via constants, operations and identities — i.e. a variety of universal algebra.

(an example of a structure that is not a universal algebra is the structure of being a field)

A related important feature is that you also have equalizers; if $f,g$ are both ring homomorphisms $R \to S$, then the subset $E = \{ r \in R \mid f(r) = g(r) \}$ is closed under the ring operations of $R$, and is thus a subring itself.

This also happens for any variety of universal algebra.


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