Why is $(Q[x],+, x)$ a ring but not a field?

View step-by-step help

Home

Math Problems

Algebra 1

Square roots

Full solution

Q. Why is

(Q[x],+, x)

a ring but not a field?

Identify $Q[x]$ and operations: Identify the set $Q[x]$ and the operations $+$ and $\times$ . $\newline$ $Q[x]$ is the set of all polynomials with rational coefficients. The operations $+$ and $\times$ refer to polynomial addition and multiplication, respectively.
Check ring properties: Check the ring properties for $(\mathbb{Q}[x],+,\times)$ . A ring is a set equipped with two binary operations (here, + and $\times$ ) satisfying the following properties: $1$ . The set is closed under addition and multiplication. $2$ . Addition is associative and commutative, and there is an additive identity ( $0$ ) and additive inverses ( $-p(x)$ for any $p(x)$ in $\mathbb{Q}[x]$ ). $3$ . Multiplication is associative and has a multiplicative identity ( $1$ ). $4$ . Multiplication is distributive over addition. $\mathbb{Q}[x]$ satisfies all these properties, so it is a ring.
Check field properties: Check the field properties for ( extbf{Q}[x],+,\times)\. A field is a ring with the additional property that every non-zero element has a multiplicative inverse in the set. For \$ extbf{Q}[x], while the constant non-zero polynomials do have multiplicative inverses in $extbf{Q}[x]$ , the non-constant polynomials do not. For example, the polynomial $x$ does not have a multiplicative inverse in $extbf{Q}[x]$ because there is no polynomial $p(x)$ in $extbf{Q}[x]$ such that $x \times p(x) = 1$ .
Conclude field status: Conclude whether $(\mathbb{Q}[x],+,\times)$ is a field. $\newline$ Since not every non-zero element in $\mathbb{Q}[x]$ has a multiplicative inverse, $(\mathbb{Q}[x],+,\times)$ does not satisfy the definition of a field.