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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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    Technically, you need $R$ to be commutative. – 2017-02-04
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    Why is that? What does technically mean? – 2017-02-04
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    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. – 2017-02-04
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    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. – 2017-02-04
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    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! – 2017-02-04
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    For the record: there is a typo in my second comment: it should have been "$ba=1$" – 2017-02-04

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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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    Not the direction I was asking about. – 2017-02-04
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    @elfeck: I'm sorry. Here the direction you asked (I was confuse because this direction is more obvious) – 2017-02-04
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    Why is $a^k = 1$ for some $k$? Especially in rings with $\text{char} = 0$? – 2017-02-04
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    if $b\in (a)$, then there is $k$ s.t. $b=a^k$. If $char(R)=0$, it doesn't change anything... @elfeck – 2017-02-04
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    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. – 2017-02-04
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    @elfeck: I would happy to help you more, but I don't really understand your question. Could you be more precise ? For a ring of Char(R)=0, the only unit are $1$ and $-1$, so if Char(R)=0, then obviously $a=\pm1$, but this is not important, since we didn't use the characteristic in the proof. – 2017-02-04
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    Sorry, forget I ever brought up the characteristic thing. My question: Why is this true: For any $b \in (a): ~\exists k \in \mathbb{Z}$ s.t. $b = a^k$. – 2017-02-04
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    @elfeck: I'm indeed a little bit fast, I correct my post. – 2017-02-04
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    Regarding your edit: Why should such a $r$ exists? – 2017-02-04
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    @elfeck: What is your definition of $(a)$ ? – 2017-02-04
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    $(a) = \{ r_1as_1 + \ldots + r_nas_n ~|~ r_i, s_i \in R\}$. I see now that for commutative $R$ we have $(a) = aR$ and it is indeed trival. Comments to my original question point out problems for non-commutative $R$ so that might be where my confusion is coming from. – 2017-02-04