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Do the non-units in a commutative ring form an ideal?

The following are my thoughts on this. Have I made any incorrect assumptions?

Let $R$ be a commutative ring. Let $a, b \in N$ with $N$ being the set of non-units in $R$. We must show the following to prove $N$ is an ideal -

  1. $0 \in N$
  2. $a + b \in N$
  3. $-a \in N$
  4. $ar, ra \in N \ \forall r \in R$

1. $0 \in N$

I.e. $0$ a non-unit. This is true as $\nexists \ 0^{-1}$ such that $0 \cdot 0^{-1} = 1$

2. $a + b \in N$

Assume $a + b$ is a unit. Then $\exists \ g \in R$, $g \neq 0$ such that

$(a + b)g = 1$ $\implies$ $ag + bg = 1$

For this to be true either $a$ or $b$ must be $0$. Consider the case when $a = 0$. Then we have $bg = 1$. But this is a contradiction as $b$ is a non-unit. Hence $\nexists \ g \in R$ such that $(a + b)g = 1$. Similarly for when $b = 0$.

Therefore, $a + b$ is a non-unit.

3. $-a \in N$

I.e. Does there exist $(-a) \in N$ such that $a + (-a) = 0$?

Assume $-a$ is a unit. Then $\exists \ g \in R, g \neq$ 0, such that $(-a)g = 1$

Consider $a + (-a) = 0$

Multiplying both sides by $g$ we get

$(a + (-a))g = 0 \cdot g$

$ag + (-a)g = 0$

$ag + 1 = 0$

$-ag = 1$

$a(-g) = 1$

But this is a contradiction as $a$ is a non-unit. Hence $(-a)$ is a non-unit.

4. $ar, ra \in N \ \forall r \in R$

Assume $ar$ is a unit. Then $\exists \ g \in R, g \neq$ 0, such that $(ar)g = 1$

I.e. $a(gr) = 1$

But this is a contradiction as $a$ is a non-unit. Hence $ar$ is a non-unit.

So, to conclude, the non-units in a commutative ring do form an ideal. Are my workings correct?

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    $\begingroup$ $3 + (-2) = 1$. Your step to show that the sum of non-units is a non-unit is wrong. $\endgroup$ Nov 13 '13 at 12:37
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    $\begingroup$ Why are people downvoting the question? It clearly shows work, and it is a legitimate question one might ask. $\endgroup$ Nov 13 '13 at 12:43
  • $\begingroup$ Thats not true like @DanielFischer said , but over a commutative ring zero divisors form an ideal . $\endgroup$
    – Theorem
    Nov 13 '13 at 12:43
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    $\begingroup$ @Theorem That is also not true. If the ring has an idempotent $e\notin \{0,1\}$, then $1-e$ and $e$ are zero divisors, and an ideal containing them must contain $1$ also. $\endgroup$
    – rschwieb
    Nov 13 '13 at 13:49
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    $\begingroup$ @OlivierBégassat Good call: we can use this to remind people that good-faith questions with mistakes don't usually deserve downvotes. Answering such questions is a teaching and learning opportunity! (And Daniel Fischer made good use of it!) $\endgroup$
    – rschwieb
    Nov 13 '13 at 13:55
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In general, the set $N$ of non-units in a commutative ring $R$ doesn't form an ideal. The first point, $0\in N$, fails if and only if $R = \{0\}$ is the trivial ring, in which $0$ is a unit. Points 3. and 4. hold in any commutative ring, and your argument is correct.

However, in general, the sum of non-units need not be a non-unit. In $\mathbb{Z}$ for example, we have $3 + (-2) = 1 \notin N$. A commutative ring in which $N$ is an ideal is a local ring, a ring with a unique maximal ideal - that ideal is then the set of non-units. An example of a local ring is the ring of formal power series over a field, $K[[X]]$, in which the unique maximal ideal is the ideal generated by $X$.

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