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From Barnes and Mack's "An Algebraic Introduction to Mathematical Logic"

S is a commutative ring with 1. Show that the class of commutative rings R with 1_R = 1_S and which contain S as a subring is a variety (in the sense of universal algebra).

I figure the sets will need to be {+, $\cdot$, -, 0, 1, S} algebras, where everything in S has arity-0. Then, add on all the laws for a commutative ring, plus a bunch of laws of the form $(s + s', (s +_S s'))$. However, I can't seem to see how to make S actually inject into any algebra in the variety. For example, how do I avoid the case where $s = 0$ for all $s \in S$?

Duncan Ramage
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  • Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $\Bbb{Z}_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly? – Rob Arthan Mar 20 '17 at 22:49
  • Aside from the parenthetical at the end, it is a direct quotation.

    http://i.imgur.com/zF1LFIa.png

     – Duncan Ramage Mar 20 '17 at 22:57
  • Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism". – Duncan Ramage Mar 21 '17 at 02:18
  • Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $\mathbf{A} / \theta$ by the maximal congruence $\theta = { (a,b) : a, b \in A }$. – amrsa Mar 21 '17 at 09:58
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    I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism! – Pedro Sánchez Terraf Mar 21 '17 at 13:38
  • What is the problem with the case $1=0$? If $1=0$ in $S$ and $R$ then we talk about the zero ring (in which still $1_S=1_R$ holds) – FWE Aug 06 '18 at 21:16

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