How to prove a subset is an ideal

Question

I just started learning about Ring Theory today and I am having some trouble truly understanding and being able to apply certain concepts. The first concept I am having trouble understanding is an ideal. I know the formal definition:

Let $R$ be a ring. A nonempty subset $I$ of $R$ is called an ideal (only dealing with two-sided as of this point) of $R$ if:

(a) $I$ is an additive subgroup of $R$

(b) Given $r \in R$, $a\in I$, then $ra \in I$ and $ar \in I$

However, I am having trouble applying this to the first (and easiest) problem:

If $R$ is a commutative ring and $a \in R$, let $L(a) = \{x \in R | xa = 0\}$. Prove $L(a)$ is an ideal of $R$.

My first approach was to first prove $L(a)$ is an additive subgroup, then that $ar,ra \in I$ but I had trouble with the second part. I then tried to find a homomorphism with kernel $L(a)$ but also got stuck.

Could someone please walk me through the way I should approach this problem? I looked at several posts on here and I am nervous as my question almost seems too easy. Please keep in mind I am learning algebra entirely on my own so a very simple explanation would be immensely helpful.

Thank you.

Answer 1

Given that $R$ is a commutative ring, we want to show that for $a \in R$, $L(a) = \{x \in R : xa = 0\}$ is an ideal of $R$.

That it is an additive subgroup:

• Let $x, y \in L(a)$, then $xa = ya = 0$ and thus $(x + y)a = xa + ya = 0 + 0 = 0.$
• Given $x \in L(a)$, it's inverse is $-x$ since $xa = 0$ means $-xa = (-x)a = 0$ and hence $-x \in L(a)$.
• The additive identity is $0$ since $0a = 0$ and given $x \in L(a)$, $x = x + 0 = 0 + x$.

That it absorbs elements from the ring: Let $r \in R$, $i \in L(a)$, since $i \in L(a)$, $ia = 0$.

• Observe that $(ri)a = r(ia) = r\cdot 0 = 0$ and thus $ri \in L(a)$.
• $ir \in L(a)$ by the previous line since we are in a commutative ring.