Show that $I$ is an ideal of $R$ Let $R$ be a commutative ring and let  $a\in R$.

Show that $I=\{x\in R\mid ax=0\}$ is an ideal.

For all $b \in R$, $$bI=b\{x\in R\mid ax=0\}=\{bx\in R\mid a(bx)=0\} =\{xb\in R\mid a(x)=0\}=Ib\;,$$ thus I can say $I$ is an ideal of $R$, right? Thanks
 A: No, your manipulations are not valid. (And they would not establish what you need to establish, even if they were valid)
$b\{x\in R\mid ax = 0\} = \{bx\in R\mid ax=0\}$. The condition is still on $x$ and $x$ alone, not on $bx$, so your second equality is unjustified.  And you don't explain why you get to drop the $b$ in the condition when going from
$$\{ bx\in R\mid a(bx)=0\}$$
to
$$\{xb \in R\mid a(x) = 0\}.$$
Moreover: an ideal is more than just a set that absorbs multiplication: you also need to show that it is nonempty, and closed under differences (equivalently, is an additive subgroup of $R$).
So you need to show that:


*

*$I\neq\varnothing$. Exhibit an element of $R$ that is definitely in $I$. (Easy).

*If $x,y\in I$, then $x-y\in I$. (Also easy, but needs to be done).

*If $b\in R$ and $x\in I$, then $bx\in I$. Easy, but your manipulations are invalid and don't establish this.
Note that since $R$ is commutative, for any set $X\subseteq R$ (ideal or not), and any element $b$, $bX=\{bx\mid x\in X\} = Xb$; so the fact that $bI=Ib$ does not help in showing that the set is an ideal. You need to show that $bI\subseteq I$. (In a noncommutative ring, you would also need to show that $Ib\subseteq I$.)
A: Hint: analogous to the proof for vector spaces, because the map $\:x \mapsto f(x) = a x\:$ is $R$-linear, its kernel enjoys an R-linear structure, i.e. it forms an ideal of $R$ (a.k.a. an $R$-module). Indeed $f(x) = 0 = f(y)$ $\:\Rightarrow\:$ $f(x-y) = f(x)-f(y) = 0,\:$ and $\:f(rx) = r f(x) = 0\:$ for $r\in R$. Therefore we've shown $\:x,y\in I = ker f$ $\:\Rightarrow\:$ $x-y,\ rx\in I,\:$ i.e. $I$ is an ideal of $R$.
