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I was wondering if for a ring $R$ with $1$, the following is true:
The ideal $(a)$ generated by some element $a \in R$ is the whole ring ($(a) = R$) iff $a \in R^\times$.

The direction $a$ is unit $\Longrightarrow (a) = R$ is clear.
But what about the other direction? I suspect that it might also hold but can't quite prove it.

My attempt is to show that if a non-unit generates $R$, then I need to be able to combine a unit (or in particular $1$) just from $a$ which (hopefully) might not be possibe. Here I get stuck.

Can someone tell me if the implication $(a) = R \implies a \in R^\times$ is correct and help me to prove it, or give me a counterexmaple?

Thanks!

EDIT: To summarize the comments: $R$ has to be commutative otherwise my statement is not true. If $R$ commutative, then the direction I was asking about holds trivially as $(a) = aR$ and $1 \in (a)$ thus $1 = ar$ and $a \in R^\times$.

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    $\begingroup$ Technically, you need $R$ to be commutative. $\endgroup$
    – user228113
    Feb 4, 2017 at 16:16
  • $\begingroup$ Why is that? What does technically mean? $\endgroup$
    – elfeck
    Feb 4, 2017 at 16:19
  • $\begingroup$ I mean: if by $(a)$ you mean the smallest two-sided ideal of $R$ containing $a$, then it might not be true that $(a)=R\implies a\in R^*$. A basic obstruction is that $ab=1$ might not imply that there is $c$ such that $ac=1$. For instance, in the ring of linear maps $\Bbb R[x]\to\Bbb R[x]$ we have some surjective maps that have a right inverse but not a left inverse, and injective maps that do the opposite. $\endgroup$
    – user228113
    Feb 4, 2017 at 16:27
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    $\begingroup$ A less obvious obstruction is that, in a non-commutative ring, the smallest two sided ideal containing $a$ is in general strictly larger than $$\{ka\,:\,k\in R\}$$ because it must also contain elements in the form $k\cdot a\cdot h+s\cdot a\cdot t$ and the likes of it. For instance, the ring $M_n(\Bbb R)$ has exactly two two-sided ideals: $\{0\}$ and itself. That being said, this is one of many reasons why noncommutative algebra is often treated separately from commutative algebra in courses. $\endgroup$
    – user228113
    Feb 4, 2017 at 16:29
  • $\begingroup$ Okay I thing I get the idea. Also I just checked and for the particular question I need this equality for I do have a commutative $R$. Thanks for your remark! $\endgroup$
    – elfeck
    Feb 4, 2017 at 16:35

1 Answer 1

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If $(a)=R$, then $1\in (a)$. In particular, $ra=1$ for some $r\in R$, and thus $a$ is a unit.

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  • $\begingroup$ Not the direction I was asking about. $\endgroup$
    – elfeck
    Feb 4, 2017 at 16:13
  • $\begingroup$ @elfeck: I'm sorry. Here the direction you asked (I was confuse because this direction is more obvious) $\endgroup$
    – Surb
    Feb 4, 2017 at 16:14
  • $\begingroup$ Why is $a^k = 1$ for some $k$? Especially in rings with $\text{char} = 0$? $\endgroup$
    – elfeck
    Feb 4, 2017 at 16:14
  • $\begingroup$ if $b\in (a)$, then there is $k$ s.t. $b=a^k$. If $char(R)=0$, it doesn't change anything... @elfeck $\endgroup$
    – Surb
    Feb 4, 2017 at 16:18
  • $\begingroup$ Okay disregard the characteristic thing, I wanted to point out that this is not clear to me especially for infinite rings. Can you outline a quick proof? Sorry maybe I am standing on the line here. $\endgroup$
    – elfeck
    Feb 4, 2017 at 16:22

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