What are the maximal ideals of Z X?

What are the maximal ideals of Z X?

The maximal ideals of Z[x] are of the form (p, f(x)) where p is a prime number and f(x) is a polynomial in Z[x] which is irreducible modulo p. To prove this let M be a maximal ideal of Z[x].

What are the prime and maximal ideals of Z?

(1) The prime ideals of Z are (0),(2),(3),(5),…; these are all maximal except (0). (3) If A = Z[x], the polynomial ring in one variable over Z and p is a prime number, then (0), (p), (x), and (p, x) = {ap + bX|a, b ∈ A} are all prime ideals of A. Of these, only (p, x) is maximal.

Is 2Z a maximal ideal of Z?

Which are maximal? The ideal 2Z ⊂ Z is prime and maximal, so that 2Z/8Z ⊂ Z/8Z is a prime and maximal ideal. The ideals Z,4Z,8Z ⊂ Z are neither prime nor maximal, so that the ideals Z/8Z,4Z/8Z,(0) ⊂ Z/8Z are neither prime nor maximal.

Is X 2 a maximal ideal?

⟨x,2⟩ is a maximal ideal in Z[x] and since ⟨x,2⟩ is an ideal, it absorbs 2k+xg(x).

Are maximal ideals principal?

In the ring Z of integers, the maximal ideals are the principal ideals generated by a prime number. More generally, all nonzero prime ideals are maximal in a principal ideal domain.

How do you show an ideal is maximal?

The Attempt at a Solution I know that there are two ways to prove an ideal is maximal: You can show that, in the ring R, whenever J is an ideal such that M is contained by J, then M=J or J=R. Or you can show that the quotient ring R/M is a field.

Are all maximal ideals prime ideals?

In a commutative ring with unity, every maximal ideal is a prime ideal. The converse is not always true: for example, in any nonfield integral domain the zero ideal is a prime ideal which is not maximal.

How many ring Homomorphisms are there from Z to Z?

By question #22 the only possible values for φ((1, 0)) are (0, 0),(1, 0),(0, 1) and (1, 1). Similarly, the only possible values for φ((0, 1)) are these same 4 values. Thus, in total there are at most 16 possible ring homomorphisms from Z⊕Z to Z ⊕ Z. However, not all of these 16 maps are ring homomorphisms.

Are prime ideals maximal?

As with commutative rings, maximal ideals are prime, and also prime ideals contain minimal prime ideals. A ring is a prime ring if and only if the zero ideal is a prime ideal, and moreover a ring is a domain if and only if the zero ideal is a completely prime ideal.

Why is the ideal 6Z not a maximal ideal of Z?

Typically proving that an ideal is maximal involves taking another ideal B with A ⊊ B and showing B = R. Typically to show B = R we show 1 ∈ B because then r = r(1) ∈ B for any r ∈ R. Example: The ideal 6Z is not maximal in Z because 6Z ⊊ 2Z⊊Z.

Is 0 a maximal ideal in Z?

Examples. If F is a field, then the only maximal ideal is {0}. In the ring Z of integers, the maximal ideals are the principal ideals generated by a prime number. More generally, all nonzero prime ideals are maximal in a principal ideal domain.

How to prove the maximal ideals of Z?

Maximal ideals of Zx]. is a prime number and Z . To prove this let M be a Z M Z 6= (0). As Z M ) injects into x/M M Maximal ideals of Z[x]. The maximal ideals of Z[x] are of the form (p,f(x)) where p is a prime number and f(x) is a polynomial in Z[x] which is irreducible modulo p. To prove this let M be a maximal ideal of Z[x].

Which is the maximal ideal of the ring?

(3) If A= Z[x], the polynomial ring in one variable over Z and pis a prime number, then (0), (p), (x), and (p,x) = {ap+ bX|a,b∈ A} are all prime ideals of A. Of these, only (p,x) is maximal. (4) If Ais a ring of R-valued functions on a set for any integral domain Rthen I= {f∈ A|f(x) = 0} is prime. Exercise. What are the prime ideals of R[X]?

When is the maximal ideal a prime ideal?

The only time these ideals are prime ideals are when n = p is prime and N = pZ (hence the term “prime ideal”). By Example 27.10, these are exactly the maximal ideals in R = Z. This illustrates Corollary 27.16 in that the maximal ideals pZ are all prime ideals. V.27 Prime and Maximal Ideals 5 Note.

Is the principal ideal of f ( x ) maximal?

No principal ideal (f(x)) is maximal. If f(x) is an integer n 6= ±1, then (n,x) is a bigger ideal that is not the whole ring. If f(x) has positive degree, then take any prime number p that does not divide the leading coefficient of f(x).