Showing (2) + (x) is not a principal ideal in Z[x]

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VincentCheung
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Hi there, I am struggling to understand this question and am unconvinced with my approach. So a principal ideal is one where each element can be written as a 'multiple' of the other.

So (2) + (x) = { a2 + bx | a , b Image[x] }, it seems too simple to just plug in values for a and b to get two elements of the ideal and then show they are not multiples of each other.. Also I am unsure on the second part of the question as
Image is not a field(i think).
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Noble.
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Well the ideal generated by 2 and x is equivalent to \{2f(x) + x g(x) \ : \ f,g \in \mathbb{Z}[x] \}

The key thing to notice is that any polynomial in this ideal will have even constant term (because once you multiply g(x) by x, you no longer have a constant term and so the constant of any polynomial in the ideal comes from the 2f(x) which will clearly be even).

Now, suppose that this ideal is principal. So there exists some h(x) \in \mathbb{Z}[x] such that (2,x) = (h(x)). You quite quickly get a contradiction to this, suppose h(x) were a constant polynomial - what's wrong with this? Then suppose it's a non-constant polynomial of degree at least one - you fail to include which elements of (2,x)?

\mathbb{Z}[x] is not a field, it's a ring (because inverses to elements do not belong to the set). To show F[x_1,\ldots,x_n] is not a PID you want to use induction. You've already done the base step by showing (2,x) is not principal, you just need to adapt the argument slightly to do the inductive step.
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VincentCheung
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Ah that makes sense, so assume Image is a constant polynomial is wrong because we have Image which has at least degree 1 so Image also does. So it's established that Image cannot be a constant polynomial. Then if Image is at least degree 1, I'm unsure how to explain it but say Image is equal to x + 2, then it is missing the x + 4 term?

For the second part, can I use the same ideal for Image since it is not the inverse elements that have determined the (2,x) is not principal. Then using induction, since the field in the first element is not a PID neither can any field in n elements?
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