Principal ideal Puns

Cual seria tu trabajo ideal, ganas lo suficiente para vivir normal, nada de lujos, no hay fama, no hay aumento y siendo la labor la satisfacción principal

Hey! I have this question (please click the link to see it, reddit crops half of it off) here that I’m stuck on.

So far, I’ve divided the two polynomials and got the answer as (x-1) + 2/(x+1) but I’m not sure where to go from here.

Any help is really appreciated, thank you :)

Also calculate in the ring Z[i] the lcm and gcd of 14+2i and 21+26i

This is a property which can be easily seen for the ring of integers of low degree fields, but for higher degree fields this turns into what I see as the least obvious equality of all time. This basically says that the product of the roots of the minimal polynomial of the ideal generator is equal to the determinant of the map sending the ring onto the ideal. Another way to kinda rephrase this is how does the determinant relate to the roots of a polynomial?

Let I = ( f (x) ) be a principal ideal in Z [ x ] . I understand that if f(x) = a , a constant, then the ideal ( a , x ) contains ( a ) , so ( a ) is not maximal. However, consider the case where f(x) has positive degree. A source I found online said to consider a prime p not dividing the leading term of f(x) . Then the ideal (p, f(x) ) contains ( f(x) ) . Why do we need to consider p not dividing the leading term of f(x) ? Source here: https://math.stackexchange.com/questions/2120064/why-principal-ideals-in-zx-are-not-maximal

If not, in what circumstance does it and why? Also, the containment from the right to the left follows directly from distributivity, right?

The title says it all: I need assistance proving that the ideal <2,1+√(-5)> is not principal in Z[√(-5)].

I've started off by supposing that it is principal and generated by x+y√(-5). Then, there exist A=a+a'√(-5) and B=b+b'√(-5) in Z[√(-5)] such that

2 = A(x+y√(-5)) and 1+√(-5) = B(x+y√(-5)).

I've played around with expanding the terms and rearranging them, but I can't see anything that forces a contradiction. Can anyone point me in the right direction? All help is greatly appreciated. Thanks.

I am still fairly new to Ring theory, so I could use a little direction on this one. I'd rather just get advice than the complete solution, as I feel like this shouldn't be too difficult.

Here Z[x] is the ring of polynomials over the integers.

I think I understand the proof that the maximal ideals of Z[x] are generated by an irreducible polynomial and a prime, but I am having trouble determining which, if any, of these are principal ideals.

Other than the trivial case, how should I be looking at this? I was thinking that ideals generated by an irreducible polynomial whose coefficients are relatively prime might work, but of course (x) is not maximal, for example.

Any suggestions?

Cual seria tu trabajo ideal, ganas lo suficiente para vivir normal, nada de lujos, no hay fama, no hay aumento y siendo la labor la satisfacción principal

Let I = ( f (x) ) be a principal ideal in Z [ x ] . I understand that if f(x) = a , a constant, then the ideal ( a , x ) contains ( a ) , so ( a ) is not maximal. However, consider the case where f(x) has positive degree. A source I found online said to consider a prime p not dividing the leading term of f(x) . Then the ideal (p, f(x) ) contains ( f(x) ) . Why do we need to consider p not dividing the leading term of f(x) ? Source here: https://math.stackexchange.com/questions/2120064/why-principal-ideals-in-zx-are-not-maximal[1]

A principal ideal ring is a ring with the property that every ideal has the form <a>. Show that the homomorphic image of a principal ideal ring is a principal ideal ring.