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Chapter 17: Problem 20

Let \((R,+, \cdot)\) be a ring. If \(I\) is an ideal of \(R\), prove that \(I[x]\), the set of all polynomials in the indeterminate \(x\) with coefficients in \(I\), is an ideal in \(R[x]\).

Short Answer

Expert verified
Given that \(I\) is an ideal in \(R\), \(I[x]\) is an ideal in \(R[x]\). This is based on the fact that \(I[x]\) is closed under both addition and multiplication by any arbitrary element in \(R[x]\), tracing back to the corresponding closure properties of \(I\) in \(R\).

Step by step solution

01

Defining the terms

Start by stating the definitions. A ring is a set \(R\) equipped with two binary operations \(+\) and \(\cdot\) that satisfy certain properties. An ideal \(I\) of a ring \(R\) is a subset that is closed under both addition and multiplication by any element of \(R\). The polynomial ring \(R[x]\) consists of all polynomials whose coefficients belong to \(R\). \(I[x]\) represents the set of all polynomials with coefficients in \(I\).
02

Show \(I[x]\) is closed under addition.

Consider two polynomials \(f(x), g(x) \in I[x]\). Their sum, \(f(x) + g(x)\), is also a polynomial, and each term's coefficient is the sum of the corresponding coefficients in \(f(x)\) and \(g(x)\). Since ideal \(I\) is closed under addition, all these coefficients remain in \(I\), so \(f(x) + g(x) \in I[x]\). Hence, \(I[x]\) is closed under addition.
03

Show \(I[x]\) is closed under multiplication by any element in \(R[x]\).

Let \(f(x) \in I[x]\) and \(h(x) \in R[x]\). The product \(h(x)\cdot f(x)\) has its terms arising from products of terms of \(h(x)\) and \(f(x)\). The coefficients of these terms are products of a coefficient from \(h(x)\) (an element of \(R\)) and a coefficient from \(f(x)\) (an element of \(I\)). As \(I\) is an ideal in \(R\), these products remain in \(I\). Therefore, \(h(x)\cdot f(x) \in I[x]\), proving that \(I[x]\) is closed under multiplication by an arbitrary element in \(R[x]\).

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