Chapter 8: Problem 15
Let \(R\) be any ring and \(I\) any ideal in \(R\). Show that \(I[x]\) is an ideal in \(R[x]\).
Short Answer
Expert verified
The set \( I[x] \) is an ideal of \( R[x] \) because it is closed under addition and multiplication by any element of \( R[x] \).
Step by step solution
01
Define Polynomial Rings and Ideals
In the ring of polynomials \( R[x] \), elements are of the form \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), where \( a_i \in R \). An ideal \( I \) in \( R \) is a subset where for any elements \( a, b \in I \) and \( r \in R \), \( a - b \in I \) and \( ra \in I \). We want to show that \( I[x] \), consisting of polynomials with coefficients from \( I \), is an ideal of \( R[x] \).
02
Show Closure under Addition
To prove \( I[x] \) is an ideal, first check closure under addition. Take any two polynomials \( f(x), g(x) \in I[x] \). These polynomials have coefficients from \( I \). Their sum, \( (f+g)(x) \), also has coefficients that are each the sum of elements from \( I \), thus remaining in \( I[x] \). Hence, \( f(x) + g(x) \in I[x] \).
03
Show Closure Under Multiplication by R[x]
Next, we must show \( I[x] \) is closed under multiplication by any polynomial \( r(x) \in R[x] \). Let \( f(x) \in I[x] \) and \( r(x) = b_m x^m + \ldots + b_1 x + b_0\). The product \( r(x)f(x) \) results in a polynomial where each coefficient is a sum of products of the form \( b_i a_j \), where \( a_j \in I \) and \( b_i \in R \). Since \( a_j \in I \) and \( I \) is an ideal in \( R \), \( b_i a_j \in I \). Thus, each coefficient of \( r(x)f(x) \) is in \( I \), meaning \( r(x)f(x) \in I[x] \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Rings
Polynomial rings are a fascinating aspect of ring theory, dealing with expressions called polynomials. Each polynomial in a ring of polynomials, denoted as \( R[x] \), has coefficients from an underlying ring \( R \).
For example, a typical polynomial in \( R[x] \) looks like this:
Polynomial rings form the basis for more advanced studies, providing a robust framework for solving equations, analyzing functions, and understanding more complex algebraic structures.
For example, a typical polynomial in \( R[x] \) looks like this:
- \( a_n x^n + a_{n-1} x^{n-1} + \,\ldots\, + a_1 x + a_0 \)
Polynomial rings form the basis for more advanced studies, providing a robust framework for solving equations, analyzing functions, and understanding more complex algebraic structures.
Closure under Addition
Closure under addition is a key property that ensures that when you add two elements from a set, the result still belongs to that set. In this context, we're dealing with polynomials within the ideal \( I[x] \) which have their coefficients drawn from an ideal \( I \).
When you pick any two polynomials \( f(x) \) and \( g(x) \) from \( I[x] \), you will find that:
When you pick any two polynomials \( f(x) \) and \( g(x) \) from \( I[x] \), you will find that:
- Each coefficient of \( (f + g)(x) \) is derived as the sum of corresponding coefficients from \( f(x) \) and \( g(x) \).
- If these coefficients were individually from \( I \), their sum will remain in \( I \).
Closure under Multiplication
Closure under multiplication is crucial for establishing that a subset behaves like a ring. In the case of \( I[x] \), it means that multiplying any polynomial in \( I[x] \) with any polynomial from \( R[x] \) results in a polynomial still in \( I[x] \).
Consider a polynomial \( f(x) \) in \( I[x] \) and another polynomial \( r(x) \) in \( R[x] \). When you multiply these two:
Consider a polynomial \( f(x) \) in \( I[x] \) and another polynomial \( r(x) \) in \( R[x] \). When you multiply these two:
- The coefficients of the resulting polynomial are formed by multiplying each coefficient of \( f(x) \) by each coefficient of \( r(x) \).
- Each term in the polynomial \( f(x) \) has coefficients from \( I \), and since \( I \) is an ideal, these products remain in \( I \).