Question

Let be a commutative ring with identity and $a\in R$. If .$a$³ = 0 _R . Show that 1 _R+ $ax$is a unit in role="math" localid="1648597323532" $R\left[x\right]$. [Hint: Consider $1+ax+a$² x²] If $a$⁴ =0_R, Show that 1_R+$ax$a unit in $R\left[x\right]$

Short answer

(a) It is proved that $1+ax$is a unit.

(b) It is proved that $1+ax$is a unit.

Step-by-step solution

Polynomial Arithmetic

If any given function $R\left[x\right]$is a ring, then the commutative, associative, and distributive laws hold such that the function $f\left(x\right)+g\left(x\right)$exists.

Polynomial Operations

Let us assume, $1+ax+a$²x², then we have:

$(1+ax)(1+ax+a$² x²) = $1-ax+a$²x²$+ax-a$²x² + a³ x³

⁼$1+a$³b³$=1+0=1$

Clearly, $1+ax$is a unit.

Hence proved, $1+ax$is a unit.

Polynomial Operations

Let us assume $1+ax+a$²x^{2 _} a³ x³, , then we have:

$(1+ax)(1-ax+a$²x ²- a³x ³) = $1-ax+a$²x²- a³ x³$+ax-$a²b² + a³b ^{3 _}a⁴x⁴

=$1-a$⁴b⁴

=$1-0=1$

Clearly, $(1+ax)$is a unit.

Hence proved, $1+ax$is a unit.