Consider the ring R=Z[1 / 2}={a / 2^n: a ∈Zn n ∈ℕ} of binary rationals. (i) Prove that R is the

Consider the ring R=Z[1 / 2}={a / 2^n: a ∈Zn n ∈ℕ} of binary...

Key Concepts

Subring Generation

This concept involves constructing a subring of a larger ring that contains a given set of elements. In general, the smallest subring containing specific elements is formed by taking all possible finite sums, differences, and products of those elements and the ring’s identity elements. This idea is crucial in understanding how R = Z[1/2] is built starting from the integers and the specific element 1/2, ensuring that the operations of the ring remain closed under the usual algebraic operations.

Units in a Ring

Units are the elements of a ring that have multiplicative inverses within the ring. Determining the units of a ring like Z[1/2] involves identifying those elements a/2^n for which there exists another element b/2^m such that their product is 1. This concept is central to understanding the multiplicative structure of a ring and plays an important role when investigating factorization properties.

Unique Factorization Domain (UFD)

A Unique Factorization Domain is a ring in which every nonzero element can be factored uniquely (up to order and units) into irreducible elements. This property guarantees that notions such as greatest common divisors are well-defined up to associates. Recognizing that R is a UFD allows one to use factorization techniques to analyze elements and polynomials within the ring.

Normal Form and GCD Function

In the context of a UFD, a normal form provides a canonical way to represent each element, often by uniquely decomposing it into irreducible factors and unit components. This normalization is used to define a meaningful greatest common divisor (gcd) function that is unique up to multiplication by a unit. For the ring of binary rationals, the normal form typically involves writing elements as an integer numerator over a power of 2, which facilitates the computation of gcds.

Content and Primitive Part of a Polynomial

The content of a polynomial over a ring is the greatest common divisor of its coefficients, and the polynomial is said to be primitive if its content is a unit in that ring. Separating a polynomial into its content and its primitive part is essential for various factorization theorems and for understanding divisibility properties. This concept is applied differently depending on whether the ring in question is Z, a UFD like Z[1/2], or a field like Q.

Transcript

00:01 We are going to show that if a rational number alpha satisfies a monic polynomial with integer coefficients, that is, if alpha is a rational number n -alpha to the n plus c -n minus 1, alpha to the n -1, plus up to c2 alpha -square plus c1 -c1 -c1 -c2 -a -fffla -f plus c1 -1 -c1 -1, up to c n minus 1 in the integer numbers, then alpha must be an integer.

00:41 That is, we have a rational number alpha, which is a root or a zero of a polynomial with integer coefficients and the coefficient of the power n of the variable is one, that is somonic polynomial.

01:05 Okay, so let's say if alpha is a rational number, we can write alpha as a quotient m, let's say, r or res, where s different from zero, or first, let's say, r and s are integer numbers.

01:49 S is not zero and in fact we can consider r and s relative primes that is the only divisor of r &s will be one or negative one can be considered to be relative primes so the only common divisor of r &s are one or negative one so having that, we can put this, remember this property here that we can consider numbers as relative primes is basically the exercise 43.

03:19 Because if we, if they are not relative primes, we can also, there will be a common divisor other than one or negative one.

03:27 We take it out as a common factor and we simplify the fraction.

03:30 So we can always do that.

03:35 Putting this alpha, r over s, into this expression, here we get r over s to the n plus c n minus 1, r over s to the n minus 1 of 2, c2, r over s square plus c1 r over s plus c0 equals 0, that is, r to the n over s to the n plus c n minus 1 r to the n minus 1 over s to the n minus 1 of 2 c2 r square over s plus c1 r over s plus c0 equal 0 and we can multiply both sides by s to the n so we get r to the n plus c n minus 1 r to the n minus 1 s up to c2 r square s to the n minus 2 plus c1 r s to the n minus 1 plus c0 s to the n okay we have this and we can now say that r to the n is equal to negative c n n minus 1 r to the n 1 s plus up to c2 r square s to the n minus 1 minus 2 minus 2 sorry plus sorry many mistake here because basically what i'm doing here in this step is to pass all the terms to the right except this 1 so it's negative for all the terms minus minus c1 r s to the n minus 1 minus 0 s to the n.

05:52 Now all these terms has a common factor s.

05:57 In fact, the lowest exponent we have for s will be s to the 1.

06:06 So we can take out a common factor s and we get s times negative c, n minus 1, r to the n minus 1.

06:19 If we write the following term, which we didn't write here, but can write here right now, will be cn minus 2, r to the n minus 2, and because we took out of common factor s, will be s, because it originally will be s square.

06:42 So we get this minus up to c2, r square, s to the n square, s to the n, minus 3.

06:59 Okay, let's put that a little bit right here.

07:14 Let's take this and put it here.

07:17 And let's take this and put it like it and take up this and put it here...

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