Why is $( extbf{Z}[x],+, imes)$ a ring bot not a field?

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Classify quadrilaterals I

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Q. Why is

( extbf{Z}[x],+, imes)

a ring bot not a field?

Check Abelian Group: To determine if $( ext{Z}[x],+,\times)$ is a ring, we need to check if it satisfies the ring axioms: $\newline$ $1$ . $( ext{Z}[x],+)$ is an abelian group. $\newline$ $2$ . $( ext{Z}[x],\times)$ is a semigroup. $\newline$ $3$ . Multiplication is distributive with respect to addition.
Verify Semigroup: First, we check if $( ext{Z}[x],+)$ is an abelian group. This means it must have an identity element, every element must have an inverse, it must be closed under addition, and addition must be associative and commutative. $\newline$ - The identity element for addition is the zero polynomial. $\newline$ - Every polynomial has an additive inverse, which is the polynomial with all coefficients negated. $\newline$ - The sum of any two polynomials in $ext{Z}[x]$ is another polynomial in $ext{Z}[x]$ , so it is closed under addition. $\newline$ - Polynomial addition is associative and commutative. $\newline$ Therefore, $( ext{Z}[x],+)$ is an abelian group.
Distributive Property: Next, we check if $( ext{Z}[x],x)$ is a semigroup, which means it must be closed under multiplication and multiplication must be associative. $\newline$ - The product of any two polynomials in $ext{Z}[x]$ is another polynomial in $ext{Z}[x]$ , so it is closed under multiplication. $\newline$ - Polynomial multiplication is associative. $\newline$ Therefore, $( ext{Z}[x],x)$ is a semigroup.
Ring Axioms Satisfied: Finally, we check if multiplication is distributive with respect to addition in $Z[x]$ . For any polynomials $a$ , $b$ , and $c$ in $Z[x]$ , the distributive property states that $a*(b+c) = a*b + a*c$ and $(b+c)*a = b*a + c*a$ . This property holds for polynomial multiplication. $\newline$ Therefore, multiplication is distributive with respect to addition in $Z[x]$ .
Check Field Criteria: Since $( ext{Z}[x],+, imes)$ satisfies all the ring axioms, it is indeed a ring.
Multiplicative Inverse Absent: Now, to determine why $( extbf{Z}[x],+,\times)$ is not a field, we need to check if every non-zero element has a multiplicative inverse in $extbf{Z}[x]$ . A field requires that every non-zero element has a multiplicative inverse within the set.
Not a Field: Consider the polynomial $f(x) = 2$ in $\mathbb{Z}[x]$ . This is a non-zero element. To have a multiplicative inverse, there must exist a polynomial $g(x)$ in $\mathbb{Z}[x]$ such that $f(x)\cdot g(x) = 1$ . However, there is no polynomial with integer coefficients that can be multiplied by $2$ to yield $1$ , since $\frac{1}{2}$ is not an integer. Therefore, not every non-zero element in $\mathbb{Z}[x]$ has a multiplicative inverse in $\mathbb{Z}[x]$ .
Not a Field: Consider the polynomial $f(x) = 2$ in $\mathbb{Z}[x]$ . This is a non-zero element. To have a multiplicative inverse, there must exist a polynomial $g(x)$ in $\mathbb{Z}[x]$ such that $f(x)\cdot g(x) = 1$ . However, there is no polynomial with integer coefficients that can be multiplied by $2$ to yield $1$ , since $\frac{1}{2}$ is not an integer. Therefore, not every non-zero element in $\mathbb{Z}[x]$ has a multiplicative inverse in $\mathbb{Z}[x]$ . Since not every non-zero element in $\mathbb{Z}[x]$ has a multiplicative inverse, $\mathbb{Z}[x]$ $1$ does not satisfy the definition of a field.