4

My book defines an ideal to be a subring of R such that $xr \in I$ and $rx \in I$, whenever $r \in R$ and $x \in I$. However, by definition any subring of a ring with unity must contain unity. So it follows that in any ring with unity, the only ideal is itself.

Surely my reasoning is incorrect, but I can't figure out where I'm going wrong.

user61924
  • 41
  • 1
    Are we sure the author requires rings to have unity in that book? – yearning4pi Feb 12 '13 at 01:47
  • No, but a subring of a ring with unity must have unity, right? – user61924 Feb 12 '13 at 01:50
  • I forgot that it is a convention among some authors to allow subrings to have a different multiplicative identity. This does not clear it up though. – yearning4pi Feb 12 '13 at 02:11
  • 4
    It depends on definitions. If "ring" doesn't include unity, a subring of a ring with unity doesn't have to have a unity. – vonbrand Feb 12 '13 at 02:31
  • 1
    "A subring of a ring with unity is supposed to have a unity" (and, similarly, "a module over a ring with unity is supposed to have the unity act trivially on it") is an incantation used by some authors who are too lazy to make up their mind and follow a consistent choice of notations. I am not surprised that they don't always make good on the promise. – darij grinberg Aug 24 '15 at 00:09

2 Answers2

2

An ideal needn't contain the unit. Consider the ring $\mathbb{Z}$ (which has the unit $1$) and the set $2\mathbb{Z} = \{ 2z \in \mathbb{Z} : z \in \mathbb{Z}\}$. It is easy to check that $2\mathbb{Z}$ is a subring.

Then for all $z\in \mathbb{Z}$ and all $z' \in 2\mathbb{Z}$, we have that $zz' \in 2\mathbb{Z}$, as we wish.

arturo
  • 51
2

If your book is defining an ideal to be a subring of $R$ with the multiplication property you described above, then it is almost certain that a subring of a ring with unity need not contain unity, because as you pointed out, this would imply that the only ideal of a unital ring is the entire ring. This definition would cause us a lot of trouble for many reasons: for example, the statement that every ideal is the kernel of some ring homomorphism and vice versa would fail massively. To resolve this dilemma, you could say that an ideal is simply an additive subgroup of the ring with the "absorption" multiplication property you described above, or you could allow unital rings to have nonunital subrings (e.g. $n\mathbb{Z}\subseteq\mathbb{Z}$ for $n\neq\pm 1$).

Stahl
  • 23,212