5
$\begingroup$

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?

$\endgroup$
1
  • 6
    $\begingroup$ 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. $\endgroup$
    – djechlin
    Nov 1, 2016 at 23:40

2 Answers 2

40
$\begingroup$

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

$\endgroup$
0
9
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .