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

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#1
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 [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
is not a field(i think).
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5 years ago
#2
Well the ideal generated by 2 and x is equivalent to

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 such that . 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)?

is not a field, it's a ring (because inverses to elements do not belong to the set). To show 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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#3
Ah that makes sense, so assume is a constant polynomial is wrong because we have which has at least degree 1 so also does. So it's established that cannot be a constant polynomial. Then if is at least degree 1, I'm unsure how to explain it but say is equal to x + 2, then it is missing the x + 4 term?

For the second part, can I use the same ideal for 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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