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If $R$ is an integral domain and a subring $S$ has identity $1_S$, how would you show that $1_S=1_R$ (here $1_R$ is the identity of the ring $R$)? I am unsure about what an integral domain really is and how the subring comes into play here.

Bill Dubuque
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user72195
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    As currently stated, it is very hard to see what is actually being asked. – Tobias Kildetoft Apr 22 '13 at 17:38
  • Subrings inherit the unit by definition. The question should be actually phrased as follows (see https://en.wikipedia.org/wiki/Rng_(algebra)): Let $R$ be an integral domain. Let $S$ be a ring such that the underlying rng of $S$ is a subrng of the underlying rng of $R$. Then, $S$ is a subring of $R$. – Martin Brandenburg Jun 09 '21 at 02:22

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If $1_S$ is a multiplicative identity of $S$, then in particular $1_S1_S=1_S$. But $1_S=1_S1_R$. It follows that $$1_S(1_S-1_R)=0.$$ In an integral domain, if $ab=0$, then $a=0$ or $b=0$. It follows that $1_S=0$ or $I_S-1_R=0$, meaning $1_S=1_R$.

So we need to eliminate the possibility $1_S=0$. That is eliminated by the assumption that $S$ is a non-trivial subring of $R$.

Remark: The condition that a product of two non-zero objects is non-zero is a key part of the definition of integral domain. It is a property common to many familiar rings, such as the ring of integers, the ring of complex numbers, the ring of polynomials with rational coefficients, and many others.

But there are plenty of interesting commutative rings with unit which are not integral domains. For example, consider the ring whose elements are $0,1,2,\dots,11$, under addition and multiplication modulo $12$. Then $(3)(4)=0$, but neither $3$ nor $4$ is equal to $0$.

André Nicolas
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  • We can produce examples of rings $R$ and subrings $S$ where this does happen. For example, the ring $R$ could be the $2\times 2$ matrices under the usual addition and multiplication, and $S$ could be the subring of matrices whose whose entries except the one in the upper left-hand corner are $0$. Naturally, in view of the result proved above, $R$ cannot be an integral domain. – André Nicolas Apr 22 '13 at 18:05
  • @user72195: The example you give doesn't work, for one thing it is not a subring, zero element is missing, and itis not closed under addition. – André Nicolas Apr 22 '13 at 18:09
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Hint $\,$ If not, the quadratic $\rm\,x^{\color{#c00}{\large 2}}\! - x\,$ has $\,\color{#c00}3\,$ roots $\rm\,0,\:1_S,\:1_R\,$ in the domain $\rm\,R,\:$ contradiction.

Note $\,$ Generally a ring R is a domain iff every nonzero polynomial over R has no more roots than its degree. This special case has easy proof: $\rm\,x(x\!-\!1)=0\:\Rightarrow\:x=0\:$ or $\rm\:x=1,\:$ by $\rm\,R\,$ a domain.

It fails in nondomains, e.g. in $\rm\,R = \Bbb Q^2\,$ the subring $\rm\:S = \Bbb Q\times \{0\}\:$ has $\rm\:1_S\! = (1,0)\neq 1_R\! = (1,1).$

Bill Dubuque
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Math Gems
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  • Note: in the above generalization, "domains" are implicitly assumed to be commutative (a rather common convention). – Bill Dubuque Jun 09 '21 at 00:59