Proof that Algebraic Integers Form a Subring

In summary: So both sums on the right side are over a finite set - finitely many zeros of the polynomials. So they can be put into an order - e.g. with respect to the length of the sum. So let's do this in an order and we see that the elements of the subgroup are linear combinations of finitely many elements with integral coefficients. So it is a finitely generated subgroup, thus the element is an algebraic integer.In summary, the set of all algebraic integers, denoted by ##\Bbb{A}##, is a subring of the complex numbers ##\Bbb{C}##. This can be seen by considering the additive subgroups ##\Bbb{Z}
  • #1
Bashyboy
1,421
5

Homework Statement



The set ##\Bbb{A}## of all the algebraic integers is a subring of ##\Bbb{C}##

Homework Equations

The Attempt at a Solution



Here is an excerpt from my book:

"Suppose ##\alpha## an ##\beta## are algebraic integers; let ##\alpha## be the root of a monic ##f(x) \in \Bbb{Z}[x]## of degree ##n##, and let ##\beta## be a root of a monic ##g(x) \in \Bbb{Z}[x]## of degree ##m##. Now ##\Bbb{Z}[\alpha \beta]## is an additive subgroup of ##G= \langle \alpha^i \beta^j ~|~ 0 \le i < n##, ~ ##0 \le j < m \rangle##. Since ##G## a finitely generated, so is its subgroup ##\Bbb{Z}[\alpha \beta]##, and so ##\alpha \beta## is an algebraic integer. Similarly, ##\Bbb{Z}[\alpha + \beta]## is an additive subgroup of ##\langle \alpha^i \beta^j ~|~ i+j \le n+m-1 \rangle##, and so ##\alpha + \beta## is also algebraic."

I am having trouble seeing the two set inclusions, particularly because ##\Bbb{Z}[\alpha] := \{g(\alpha) ~|~ g(x) \in \Bbb{Z}[x] \}## and the degree of the polynomials in ##\Bbb{Z}[x]## is unbounded, while ##G## and the other set are built from (multivariable) polynomials of finite degree. Perhaps someone could make this more explicit. Also, what's the motivation for choosing ##n+m-1## as the upper bound for ##i+j##, other than the fact that it works?
 
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  • #2
The degrees in ##\mathbb{Z}[x]## maybe be arbitrary - I don't like unbounded here, as they are definitely finite for every element - but the algebraic expressions can be reduced. Let me first order your notation, as the double use of ##g(x)## is dirty. So let's stick with the first definition as minimal polynomial of degree ##m## with ##g(\alpha)=0##. Then for any element ##p(x) \in \mathbb{Z}[x]## we have ##p(\alpha) \equiv r(\alpha) \operatorname{modulo} g(\alpha)## with ##\deg(r) < m## by use of the Euclidean algorithm. So ##\mathbb{Z}[x]/\langle g(x) \rangle \cong \mathbb{Z}[\alpha]## has to be compared with ##G##, not the entire polynomial ring.

The minus one in ##n+m-1## comes in with the division, same as in the above with ##\deg(r) \leq m-1##.
 

Related to Proof that Algebraic Integers Form a Subring

1. What are algebraic integers?

Algebraic integers are numbers that are roots of monic polynomials with integer coefficients. In other words, they are solutions to polynomial equations with integer coefficients.

2. What is a subring?

A subring is a subset of a ring that is itself a ring, meaning it is closed under addition, subtraction, and multiplication. In other words, the elements of a subring can be added, subtracted, and multiplied with each other.

3. How do you prove that algebraic integers form a subring?

To prove that algebraic integers form a subring, we need to show that they satisfy the two properties of a subring: closure under addition and multiplication. This means that when we add or multiply two algebraic integers, the result is also an algebraic integer.

4. Why is it important to prove that algebraic integers form a subring?

Proving that algebraic integers form a subring is important because it allows us to make statements and perform calculations with these numbers in a more general and systematic way. It also allows us to define and work with more complex mathematical structures, such as algebraic number fields.

5. Can algebraic integers form a subring of any ring?

No, algebraic integers can only form a subring of a special type of ring called a Dedekind domain. This is because Dedekind domains have certain properties that are necessary for the closure under addition and multiplication of algebraic integers.

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