How do we show that these ideals are co-maximal?

Suppose we have R=R[x]R=\mathbb{R}[x] the ring of polynomials and the ideals I=(x2+1)I=(x^2+1) and J=(x+1)J=(x+1). We need to show that II and JJ are co-maximal, so we want I+J=RI + J = R.

I have absolutely no idea how to show this. For example, how do I show that 5×2+1∈I+J5x^2+1 \in I + J? We know 5×2+5∈I+J5x^2+5 \in I+J because it’s an element of II. But I don’t see how we can show that all the elements of R are in I+JI+J.

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1 Answer
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You only have to show that a unit, like 11 simply, is contained in your ideal. This can be done directly, via something like 12(x2+1)−12(x−1)(x+1)=1∈I+J\frac{1}{2}(x^2+1) – \frac{1}{2}(x-1)(x+1) = 1 \in I + J. Additionally, since R\mathbb{R} is a field R[x]\mathbb{R}[x] is a Principal Ideal Domain (and also a Unique Factorization Domain), and you use the fact that x2+1,x+1x^2+1, x+1 are coprime (since they are essentially distinct irreducibles).

  

 

How does it follow that 12\frac{1}{2} and (x−1)(x-1) are in I+JI+J?
– John
2 days ago

  

 

they don’t have to be necessarily. For y∈Iy \in I we can take any ring element a∈Ra \in R and ay∈Iay \in I as well. This is part of the definition of an ideal. This is why we only need 1∈I1 \in I since this implies any a∈Ra \in R is contained in II as 1(a)∈I1(a) \in I.
– basket
2 days ago

  

 

I forgot about the definition of ideals completely. It’s totally clear now, thank you!
– John
2 days ago