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A question I recently encountered was:

"Let $R$ be a commutative ring with one not equal to zero. Let $I$ and $J$ be ideals of $R$, such that $I + J = R$. Show that for any positive integers a,b, we have that $I^a + J^b = R$. Hint: What is $I\cap J$?"

Regarding the hint, it becomes apparent that $I \cap J = IJ$. If we have an element $ij$ of $IJ$ (with $i \in I$ and $j \in J$), then by absorption on the left and right, $ij$ is in both $I$ and $J$. Hence, it is in $I\cap J$. For the other containment, take$x \in I \cap J$. Since $I + J = R$, there is $i \in I$ and $j \in J$ such that $i + j = 1$. Multiplying $x$ by $(i + j)$, we have $x = ix + xj$ which by absorption is in $R$.

Now, I'm not really sure what to do with the hint. Can anyone give me a further hint for this problem?

Robert Lewis
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ThinkConnect
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  • Those ideals are to be taken in a ring, I assume commutative. The only ideals in a field $F$ are $(0)$ and $F$, which would make the proposition rather trivial. – A.P. Jun 13 '15 at 17:00
  • I'm confused by your phrase, "ideals of a field $R$". Is $R$ a field? Fields don't have too many ideals. Does $R$ have a sublield? Can you clarify for me? Also, try $\LaTeX$; it's pretty easy. To illustrate, I'm a-gonna $\LaTeX$ify your post, yes I am. Back in a flash! Cheers! – Robert Lewis Jun 13 '15 at 17:02
  • Thank you for your observation A.P. You are correct. I edited the question to indicate that it is a commutative ring with 1 not equal to zero. – ThinkConnect Jun 13 '15 at 17:02
  • Thanks Robert! I will do that from now on. – ThinkConnect Jun 13 '15 at 17:03
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    Here you can find a brief guide on how to write math on this site. – A.P. Jun 13 '15 at 17:06
  • Hey, you beat me to it! Way to go! Doesn't it look good now! Like my main squeeze with a new 'do! Cheers! – Robert Lewis Jun 13 '15 at 17:08
  • I did catch a few things. One good way to learn more $\LaTeX$ is to check out the raw text of posts using "edit". When I see $\LaTeX$ I want to learn, that's what I do. Cheers! – Robert Lewis Jun 13 '15 at 17:16
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    Thanks Robert and A.P.! I'm new to this site, and you guys are making this very easy for me. – ThinkConnect Jun 13 '15 at 17:18
  • Well, I'm happy to do so! By the way, nice question, *endorsed!* – Robert Lewis Jun 13 '15 at 17:26

2 Answers2

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Hint : $I+J=R \Rightarrow i+j=1(i\in I,j\in J)\Rightarrow 1=(i+j)^{a+b}=\Sigma_{k=0}^{a+b} {{a+b}\choose k}i^kj^{a+b-k}$

Mojtaba
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Hint: Consider $R^2$ in the light of what you now know.

(I am assuming your definition of a ring contains a multiplicative identity)

Mark Bennet
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