MHB Maximal Ideals - Exercise 5.1 (iii) Rotman AMA

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 5.1 Prime Ideals and Maximal Ideals ...

I need some help with Exercise 5.1, Part (iii) ... ...Exercise 5.1 reads as follows:
View attachment 5943Can someone please help me to get a start on this exercise ...

Peter

 

[Deveno]

Hint:

over a field $F$, the ring $F[x]$ is a Euclidean domain (with Euclidean function $\text{deg}$), and thus a fortiori a principal ideal domain, and also a unique factorization domain.

So maximal ideals are generated by irreducible polynomials, which *are* the prime ideals as well, since irreducibles are prime and vice-versa.

Due to the Fundamental Theorem of Algebra, part (iii) is actually easier than part (ii).

 

[Math Amateur]

Hint:

over a field $F$, the ring $F[x]$ is a Euclidean domain (with Euclidean function $\text{deg}$), and thus a fortiori a principal ideal domain, and also a unique factorization domain.

So maximal ideals are generated by irreducible polynomials, which *are* the prime ideals as well, since irreducibles are prime and vice-versa.

Due to the Fundamental Theorem of Algebra, part (iii) is actually easier than part (ii).

Thanks for the help, Deveno ... appreciate it

Not quite sure what you are saying about Part (ii) though ... again we have a PID and so again, as you say, the maximal ideals are generated by irreducible polynomials, which *are* the prime ideals as well, since irreducibles are prime and vice-versa ... so answer seems the same ...

Can you clarify ...

Peter

 

[Deveno]

$\Bbb C[x]$ has a lesser variety of irreducible polynomials that $\Bbb R[x]$ does. For example, $x^2 + 1 = (x + i)(x - i)$ is reducible in $\Bbb C[x]$, but is irreducible in $\Bbb R[x]$.

This is because $\Bbb C$ is algebraically closed, and $\Bbb R$ is not.