# A ring in which the two operations are equal is {0}

Let R be a ring in which the two operations are equal, i.e., $a + b = ab \mbox{ }\forall a,b \in R$. Prove that $R = \{0 \}$.

I tried to prove that $R \subset \{0 \}$ and $\{0 \} \subset R$. For the second inclusion, we have $0 + 0 = 0 = 0 \cdot 0$. So $\{0 \} \subset R$. However, I can't figure out a way of showing that $R \subset \{0 \}$.

Any tips?

• The proof that $\{0\} \subset R$ can't possibly anything other than " $0 \in R$ by definition of $0$ and $R$". If you wrote anything else you did it wrong. Nov 1, 2016 at 23:40

For any $$a\in R$$, $$a=a+0=a\cdot 0=0$$.

Although the question has already been answered pretty accurately, I would like to detail the typical reasoning used in this case.

What you want to prove is that $R \subset \{0 \}$.

What you should do, is try to prove that every element of $R$ is also an element of $\{0\}$.

As wrote User1006, the way to achieve this is:

Let $x\in R.$ \begin{align}x+0 &= x\cdot0 \\ x\cdot0 &= 0~~~~~\textrm{ by definition of ring}\end{align} (This line is not that trivial) \begin{align} ~~x+0 &= 0\\ x&= 0\,.\end{align}

$x$ is any element of $R.$

Hence, $\forall x \in R, x \in \{0\}.$

$$~~ R \subset \{0\}.$$

This is basically what User1006 wrote, but every time you come across such a question, this is the formality you should keep in mind.