subring Archives - Epsilonify Best solutions on internet! Fri, 15 Sep 2023 23:31:11 +0000 en-US hourly 1 https://wordpress.org/?v=6.4.5 https://www.epsilonify.com/wp-content/uploads/2022/09/cropped-E-M7-32x32.png subring Archives - Epsilonify 32 32 The set {(r,r) | r in R} is a subring of the ring RxR https://www.epsilonify.com/mathematics/ring-theory/the-set-of-two-dimensional-array-with-r-and-r-where-r-is-in-r-is-a-subring-of-the-ring-r-times-r/ https://www.epsilonify.com/mathematics/ring-theory/the-set-of-two-dimensional-array-with-r-and-r-where-r-is-in-r-is-a-subring-of-the-ring-r-times-r/#respond Tue, 21 Feb 2023 13:00:37 +0000 https://www.epsilonify.com/?p=1997 Let be a ring with 1. The set is a subring of the ring . Proof. First, we want to show that the set is nonempty, but that is already easy to see since . Secondly, we want to show that is closed under subtraction. Let , then Since is closed under subtraction, we get […]

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]]> Let R be a ring with 1. The set \{(r,r) | r \in R\} is a subring of the ring R \times R.

Proof. First, we want to show that the set S = \{(r,r) | r \in R\} is nonempty, but that is already easy to see since (1,1) \in S. Secondly, we want to show that S is closed under subtraction. Let (r_1,r_1),(r_2,r_2) \in S, then 
\begin{align*}
(r_1,r_1) + (-(r_2,r_2)) &= (r_1,r_1) - (r_2,r_2) \\
&= (r_1 - r_2, r_1 - r_2).
\end{align*}
Since R is closed under subtraction, we get that (r_1 - r_2) \in R. Therefore, (r_1 - r_2, r_1 - r_2) \in S. Lastly, we need to show that S is closed under multiplication. Let (r_1,r_1),(r_2,r_2) \in S. Then we get the following: 
\begin{align*}
(r_1,r_1)(r_2,r_2) = (r_1r_2,r_1r_2). 
\end{align*}
Since R is closed under multiplication, we get that r_1r_2 \in R. Therefore, (r_1r_2,r_1r_2) \in S. Conclusion, the set \{(r,r) | r \in R\} is a subring of the ring R \times R.

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