Is a field dedekind domain?

A field is a commutative ring in which there are no nontrivial proper ideals, so that any field is a Dedekind domain, however in a rather vacuous way. Some authors add the requirement that a Dedekind domain not be a field.

What is a polynomial over a ring?

In mathematics, especially in the field of algebra, a polynomial ring or polynomial algebra is a ring (which is also a commutative algebra) formed from the set of polynomials in one or more indeterminates (traditionally also called variables) with coefficients in another ring, often a field.

Is polynomial ring finitely generated?

For example, a polynomial ring R[x] is finitely generated by {1,x} as a ring, but not as a module. If A is a commutative algebra (with unity) over R, then the following two statements are equivalent: A is a finitely generated R module. A is both a finitely generated ring over R and an integral extension of R.

Is the polynomial ring free?

The polynomial ring R[X] is a free R-module with basis 1, X, X2,… .

Is Za a UFD?

The prime elements of Z are exactly the irreducible elements – the prime numbers and their negatives. Definition 4.1. 2 An integral domain R is a unique factorization domain if the following conditions hold for each element a of R that is neither zero nor a unit. Claim: Z[√−5] is not a UFD.

Is an integral domain?

In mathematics, specifically abstract algebra, an integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero. In an integral domain, every nonzero element a has the cancellation property, that is, if a ≠ 0, an equality ab = ac implies b = c.

What is polynomial ring give example?

The ring of polynomials over R is the ring R[x] consisting of all expressions of the form a0 + a1x + a2x2 + ททท, where each ai ∈ R and all but finitely many ai’s are zero. For example p(x) = x · (x − 1)ททท(x − p + 1) defines the zero function on Zp, but is not the zero polynomial (why?).

Why is a polynomial ring not a field?

Because by definition, the only polynomial that can have a negative degree is 0, which is defined to have a degree of −∞. Non-zero constants have degree 0. You then have the degree equation: deg(fg)=deg(f)+deg(g) for any polynomials f,g.

Is every ring finitely generated?

Actually, every ring can be viewed as a module over itself generated by one element, and ideals and submodules coincide; therefore it is a more general point of view if we consider the space of modules generated by k marked elements over Z[x1,…,xn] (or even any commutative ring), the case k = 1 corresponding to the …

What does it mean for a ring to be finitely generated?

A ring is an associative algebra over the integers, hence a ℤ-ring. Accordingly a finitely generated ring is a finitely generated ℤ-algebra, and similarly for finitely presented ring. For rings every finitely generated ring is already also finitely presented.

Is Z √ 7 an UFD?

And Z[√7] = {a+b√7: a and b are in Z}, Z is a ring of integers.

Why Z is UFD?

Let R be a domain in which every irreducible element is prime. Then the decom- position of an element as product of irreducibles, if it exists, is unique. In Z, being prime is the same as being irreducible. This is the essential property that allows us to prove that Z is a UFD.

Which is an example of a Dedekind domain?

De\\fnition 1 A Dedekind domain is an integral domain that has the following three properties: (i) Noetherian, (ii) Integrally closed, (iii) All non-zero prime ideals are maximal.2 Example 1 Some important examples: (a) A PID is a Dedekind domain.

Is the local domain a PID or a Dedekind ring?

But a local domain is a Dedekind ring iff it is a PID iff it is a discrete valuation ring (DVR), so the same local characterization cannot hold for PIDs: rather, one may say that the concept of a Dedekind ring is the globalization of that of a DVR. (DD1) Every nonzero proper ideal factors into primes.

Which is an integral closure of a Dedekind domain?

In general, the integral closure of a Dedekind domain in an infinite algebraic extension is a Prüfer domain; it turns out that the ring of algebraic integers is slightly more special than this: it is a Bézout domain . Let R be an integral domain with fraction field K.

Is the fractional ideal invertible in a Dedekind domain?

Now we can appreciate (DD3): in a Dedekind domain (and only in a Dedekind domain) every fractional ideal is invertible. Thus these are precisely the class of domains for which Frac ( R )/Prin ( R) forms a group, the ideal class group Cl ( R) of R.