Problem 6 - A Book of Abstract Algebra

Question

Show that if \(A\) is an integral domain, the only invertible elements in \(A[x]\) are the constant polynomials \(\pm 1\). Then show that in \(\mathbb{Z}_{4}[x]\) there are invertible polynomials of all degrees.

Step-by-Step Solution

Answer

In an integral domain, invertible polynomials are constants \(\pm 1\). In \(\mathbb{Z}_{4}[x]\), invertible polynomials exist at any degree due to lack of divisors of zero.

Step 1: Understand Invertible Elements in Polynomial Rings

An invertible (or unit) element in a ring is an element that has a multiplicative inverse within the ring. For polynomials in ring \(A[x]\), \(f(x)\) is invertible if there exists a polynomial \(g(x)\) in \(A[x]\) such that \(f(x)g(x) = 1\).

Step 2: Consider Invertibility in an Integral Domain

Suppose \(f(x) = a_nx^n + ... + a_0\) is invertible in \(A[x]\). Multiplying by an inverse \(g(x)\), we get \((a_nx^n + ... + a_0)(b_mx^m + ... + b_0) = 1\). For the constant term of the product to be 1 and since \(A\) is an integral domain, \(a_0b_0 = 1\) meaning that \(a_0\) and \(b_0\) must both be units in \(A\).

Step 3: Analyze Degrees of Polynomials in Invertibility

For \(a_nx^n\) and \(b_mx^m\) if \(n+m > 0\), terms of degree greater than 0 will appear in \(f(x)g(x)\). Since \(1\) is of degree 0, no such terms can appear, forcing \(n+m=0\) meaning \(n=m=0\). Therefore, both \(f(x) = a_0\) and \(g(x) = b_0\) are constant polynomials.

Step 4: Conclude for Integral Domain

Thus, the only invertible polynomials in \(A[x]\) are the constant polynomials that are units in \(A\). For an integral domain like \(A\), the units are \(\pm 1\), hence invertible polynomials are \(\pm 1\) only.

Step 5: Consider Nature of \(\mathbb{Z}_{4}\)

In \(\mathbb{Z}_{4}\), more elements have inverses because \(\mathbb{Z}_4\) is not an integral domain (e.g., 2 is not invertible, and 2 \(\cdot\) 2 = 0). Elements such as 1 and 3 are invertible since 1 \(\cdot\) 1 = 1 and 3 \(\cdot\) 3 = 9 \(\equiv\) 1 (mod 4).

Step 6: Show Invertible Polynomials in \(\mathbb{Z}_{4}[x]\)

Consider polynomials like \(a_0 + a_1x\) with \(a_i\in \mathbb{Z}_4\) being invertible if their constant term is a unit in \(\mathbb{Z}_4\). Let \(f(x) = 3 + 2x\) which is invertible with \(g(x) = 3 - 2x\) since \((3 + 2x)(3 - 2x) = 9 - 4x^2 \equiv 1 (\mod 4)\).

Step 7: Demonstrate Higher Degree Inverses

In \(\mathbb{Z}_{4}[x]\), polynomials like \(3 + 2x + 3x^2\) have inverses like \(3 - 2x + 3x^2\) and checking their product shows invertibility. Such constructions can be systematically done for any degree by adjusting coefficients to ensure the polynomial product \(\equiv 1 (\mod 4)\).

Step 8: Conclude for \(\mathbb{Z}_{4}[x]\)

Therefore, \(\mathbb{Z}_{4}[x]\) allows invertible polynomials of any degree due to its properties allowing non-zero divisors to have inverse polynomials, unlike an integral domain.

Key Concepts

Invertible Elements

When we talk about invertible elements in mathematics, particularly in ring theory, we refer to elements that have a multiplicative inverse within the same structure. In simpler terms, if you have an element \( a \) in a ring \( R \), then \( a \) is invertible if there exists another element \( b \) in \( R \) such that \( ab = ba = 1 \), where 1 is the multiplicative identity of the ring.For example:

• In the ring of integers \( \mathbb{Z} \), the only invertible elements (called units) are \( 1 \) and \( -1 \), because \( 1 \times 1 = 1 \) and \( (-1) \times (-1) = 1 \).
• In the ring of integers modulo 4, \( \mathbb{Z}_4 \), the invertible elements are 1 and 3. This is because mod 4 equivalencies, \( 1 \times 1 = 1 \) and \( 3 \times 3 = 9 \equiv 1 (\mod 4) \).

Understanding invertible elements is crucial when exploring more complex algebraic structures like polynomial rings and modular arithmetic. It lays the foundation for solving equations and constructing mathematical proofs.

Polynomial Rings

Polynomial rings extend the concept of integers to polynomials. If \( A \) is a ring, then the polynomial ring \( A[x] \) involves polynomials with coefficients from \( A \). This structure allows us to perform operations like addition, subtraction, and multiplication among polynomials.Polynomials in \( A[x] \) are expressed as:

• \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \)

where \( a_i \) are elements from \( A \) and \( x \) represents the variable of the polynomial.A particularly interesting property of polynomial rings is that they can carry over the characteristic traits of the underlying ring \( A \). For instance, if \( A \) is an integral domain, meaning it has no zero-divisors, then \( A[x] \) will also inherit this property. This affects how we perceive invertibility and units within polynomial rings.In an integral domain \( A \), the invertible elements in the polynomial ring \( A[x] \) are shown to be constant polynomials such as \( \pm 1 \). This is because any polynomial with a degree higher than zero creates additional degrees when multiplied by another polynomial, which cannot yield a constant term like 1.

Units in a Ring

Units within a ring are a subset of invertible elements that hold special significance. They possess a multiplicative inverse, meaning for any unit \( u \) in ring \( R \), there is another element \( v \) such that \( uv = vu = 1 \).In integral domains such as \( \mathbb{Z} \) (the set of all integers), units are only \( 1 \) and \( -1\). This is because:

• Any non-zero element multiplied by one or its inverse (negative counterpart) yields the multiplicative identity, which is 1.
• Moreover, a great trait of units is that they do not have divisors of zero, ensuring the preservation of the non-zero product property.

When considering units in polynomial rings, the situation becomes slightly more nuanced. If we move to \( A[x] \) where \( A \) is an integral domain, the units are again the constant polynomials \( \pm 1 \). This arises because higher degree polynomial products would introduce higher degree components, precluding them from being equated to the multiplicative identity, which is a constant.Understanding the nature of units in rings helps us gauge the overall structure and predict properties or solutions feasible within that mathematical framework.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after reaching a certain value, known as the modulus. This concept is akin to how clocks operate—once you pass 12, you start back at 1.Applying this to rings, consider the structure \( \mathbb{Z}_n \), which consists of integers from 0 to \( n-1 \) with operations defined by taking the remainder when divided by \( n \). Over \( \mathbb{Z}_n \):

• 1 and any number co-prime to \( n \) (where gcd=1) can have inverses.
• For an element to have an inverse under modulo \( n \), it should be a unit, like 1 and 3 in \( \mathbb{Z}_4 \).

This notion significantly extends when considering polynomial rings such as \( \mathbb{Z}_4[x] \), allowing us to explore inverses of polynomials of various degrees by calculating the products under modulo conditions.The exceptional flexibility of modular arithmetic illustrates why solutions may exist in seemingly contradictory settings—like in \( \mathbb{Z}_4[x] \) where polynomials of any degree can be invertible. Such operations are vital in cryptography, computer science, and number theory, providing a robust toolset for understanding and manipulating numeric structures.