18
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ Take the zero subring. $\endgroup$ – Jakob Werner Jun 25 '14 at 11:58
  • $\begingroup$ What could possibly prohibit this - there is no dependency between the two having unit elements. Unit elements exist independently of this relationship. $\endgroup$ – The Messiah Jun 25 '14 at 14:23
  • $\begingroup$ www-math.mit.edu/~poonen/papers/ring.pdf $\endgroup$ – Jeff Jun 26 '14 at 12:43
27
$\begingroup$

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.

$\endgroup$
16
$\begingroup$

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.

$\endgroup$
  • 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. $\endgroup$ – martini Jun 25 '14 at 11:20
  • $\begingroup$ @martini: Thank you; corrected. $\endgroup$ – Jakub Konieczny Jun 25 '14 at 12:02
4
$\begingroup$

$\mathbb{Z}/(12)$ 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 $9$!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.