Exercise 3.8

Let $R$ be a ring. Prove that $(2 (1_{R})) \cdot (3 (1_{R})) = 6 (1_{R})$ . More generally, give an argument showing that $(m 1_{R}) \cdot (n 1_{R}) = (mn) 1_{R}$ . (Show that for all $a \in R, (m 1_{R}) \cdot a = ma$ . You may assume that $m$ is nonnegative. Why?)

Answer

By direct computation, we have
$2 (1_{R}) \cdot (3 (1_{R})) = (1_{R} + 1_{R}) \cdot (1_{R} + 1_{R} + 1_{R}) = (1_{R} + 1_{R} + 1_{R} + 1_{R} + 1_{R} + 1_{R}) = 6 (1_{R})$
Now we will prove this in general. Let $a \in R$ and $m \in Z$ . We will prove that $m 1_{R} \cdot a = ma$ . We may assume that nonnegative. We will now proceed by induction.

Base case ( $m = 0$ ): By definition, $0 (1_{R}) \cdot a = 0_{R} \cdot a = 0 a$ . Inductive step: Suppose that $m 1_{R} \cdot a = ma$ for some $m \geq 0$ . Then,
$(m + 1) 1_{R} \cdot a = (m 1_{R} + 1_{R}) \cdot a = m 1_{R} \cdot a + 1_{R} \cdot a = ma + a = (m + 1) a$
Thus, $m 1_{R} \cdot a = ma$ for all $a \in R, m \in Z$ . Then, $(m 1_{R}) \cdot (n 1_{R}) = m (n 1_{R})$ which is $m$ groups of $n$ times of $1_{R}$ , that is $mn$ times of $1_{R}$ . Thus, $(m 1_{R}) \cdot (n 1_{R}) = (mn) 1_{R}$ . $■$

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