A ring need not have a unit element, e.g. even numbers.
Can someone give me an example of a ring which doesn't have a unit element but it's subring does have a unit element.
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6$\begingroup$ Take the zero subring. $\endgroup$– user158047Jun 25, 2014 at 11:58
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$\begingroup$ What could possibly prohibit this - there is no dependency between the two having unit elements. Unit elements exist independently of this relationship. $\endgroup$– The MessiahJun 25, 2014 at 14:23
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$\begingroup$ www-math.mit.edu/~poonen/papers/ring.pdf $\endgroup$– JeffJun 26, 2014 at 12:43
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4 Answers
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A simple example:
If $R$ and $S$ are rings, where $R$ has a unit but $S$, doesn't, then $R\times S$ doesn't have a unit, but the subring $R\times\{0\}$ does.
So, for example, you could take the product $\mathbb{Z}\times2\mathbb{Z}$ of the integers and the even integers.
answered Jun 25, 2014 at 10:53
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There is no reason why not. Even if a ring $A$ fails to have a unit, it can still have an idempotent. If you just take an idempotent, together with all its integer multiples, you get a ring with unit.
(Edited from this point. Thank you for pointing out the flaw).
For example, take $A$ to be $c_{00}$, the set of all sequences $(x_n)_{n\in \mathbb{N}}$ such that all but finitely many $x_n$ are $0$. With pointwise multiplication and addition, this is a ring. The only possible unity would be the constant sequence $x_n = 1$ that obviously does not satisfy the requirement that almost all $x_n$'s are $0$'s. However, $A$ contains idempotents: these are just the $0/1$-valued sequences. To be concrete, take the sequence $e_n$ given by $e_1 = 1$, $e_n = 0$ for $n \neq 1$. Then $e \in A$ and $e^2 = e$. It is now easy to convince yourself that $B := \{k \cdot e \ : \ k \in \mathbb{Z}\}$ is a subring of $A$, and that in fact $B$ is isomorphic to $\mathbb{Z}$.
Other examples when a similar trick would work is $l^1$ (the space of summable sequences), or $L^1(\mathbb{R}) \cap B(\mathbb{R})$, the space of bounded integrable funtcions. The previous example I proposed, $L^1(\mathbb{R})$, fails to be a ring because it is not closed under multiplication.
answered Jun 25, 2014 at 10:20
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2$\begingroup$ But this isn't a ring, as $L^1(\mathbb R)$ is not closed under pointwise multiplication ... see also user160048's answer below. $\endgroup$– martiniJun 25, 2014 at 11:20
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$\mathbb{Z}/(12)$ does have a unit element, so in that sense it is not an example, but its ideal $(3)$ has a (different of course) unit element as well, namely $9$!
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Here is another example based on matrices. Let $K$ be a field. Consider the set $R$ of matrices of type $$ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \subset K^{2,2}. $$ Supply $R$ with the usual matrix addition and multiplication. Then $R$ is a ring without unit element. In fact $$ \begin{pmatrix} 1 & b \\ 0 & 0 \end{pmatrix} $$ is a left-sided unit for all $b$, but there is no right unit. However, the set of matrices of type $$ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \subset K^{2,2}. $$ is a subring of $R$, which is even a field.
