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Let $R$ be a ring and $S_i$ be the subrings of $R$ such that $R = S_1 \cup S_2$. Prove that $R = S_1$ or $R = S_2$.

I am not exactly sure what to do here. If I want to proceed with letting $R \neq S_1$ and show that $R = S_2$, what should I do? Also, proof by contradiction isn't making sense to me in this case, since if we let $R = S_1 \cup S_2$ then $R \neq S_1$ and $R \neq S_2$, now what? I don't think it can be this trivial: Since we are given with $R = S_1 \cup S_2$ And, therefore, $R = S_1$ or $R = S_2$, which is a contradiction, can it? I am all out of clues. Any help would be much appreciated.

Edit: Ok I see that similar questions have already been asked here before, but those didn't clear up my confusions, and I don't think it'd be polite to comment or ask anything on there. Also, I think my approaches are slightly different than their's.

Jellyfish
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  • Your instinct proceeding by contradiction is a good one. Think about this: If $R\neq S_1$, then $S_1\subsetneq R$. Then we can find an element with what property? – John Brevik Apr 05 '15 at 17:30
  • Take $x\in R\setminus S_1$, and $y\in R\setminus S_2$. – vadim123 Apr 05 '15 at 17:37

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Assume $\exists s_1 \in S_1$ such that $ s_1 \not\in S_2$, and $\exists s_2 \in S_2$ such that $s_2 \not\in S_1$.

You need to show that $s_1 + s_2 \not\in S_1$ and $s_1 + s_2 \not\in S_2$ which contradicts the fact that $R = S_1 \cup S_2$.

A good intuitive picture is to imagine that $R$ is a 3d vector space over the real numbers. Let $S_1$ be a plane (2d subspace) and $S_2$ be a line (1d subspace) not contained in the plane, then if you picture a vector inside the line but not in the plane, and a vector in the plane but not in the line, then their sum will be outside both the line and the plane; so the line $\cup$ the plane is closed under addition if and only if the line $\subset$ the plane.

wlad
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    We are given $R=S_1\cup S_2$ and want to show one of the subrings was $R$ itself. So assuming $R\neq S_1\cup S_2$ doesn't do any good here. – Christoph Apr 05 '15 at 20:08
  • No. If $x + y \in S_1$ then, you can write $x + y = s_1$ for some $s_1 \in S_1$, so $y = s_1 + (-y)$, which implies $y \in S_1$ because a subspace is closed under addition, which from the way we defined $y$ must be false.. You dispatch $x + y \in S_2$ in the same way by showing that it would imply that $x \in S_2$ which we know is false from the way you defined $x$. – wlad Apr 05 '15 at 21:50
  • You edited the post after my comment. It started with "Assume $S_1\cup S_2\neq R$" before. I can read. – Christoph Jul 04 '15 at 08:44