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UPSC 2018 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Integral domains; characteristic · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Let $R$ be an integral domain with unit element. Show that any unit in $R[x]$ is a unit in $R$.

Technique

Additivity of the degree function in $R[x]$ over an integral domain; reduce a unit equation $fg=1$ to a degree equation.

Solution

Here “unit element” means $R$ has a multiplicative identity $1\ne 0$, and a unit of a ring is an element with a two-sided multiplicative inverse. We must show: if $f(x)\in R[x]$ is a unit of the polynomial ring $R[x]$, then in fact $f$ is a constant polynomial whose constant term is a unit of $R$ (so $f\in R$ and is a unit there).

Step 1 — Degree function on $R[x]$ when $R$ is a domain

For a nonzero polynomial $f=a_0+a_1x+\cdots+a_mx^m$ with $a_m\ne0$ write $\deg f=m$ and call $a_m$ the leading coefficient. The crucial fact, valid precisely because $R$ is an integral domain (no zero divisors):

Indeed the top coefficient of $fg$ is $a_mb_n$, and $a_m\ne0,\ b_n\ne0\Rightarrow a_mb_n\ne0$ since $R$ has no zero divisors. In particular $R[x]$ is itself an integral domain and the degree is additive.

Step 2 — A unit must have degree $0$

Let $f\in R[x]$ be a unit, so there exists $g\in R[x]$ with

Neither $f$ nor $g$ is the zero polynomial (their product is $1\ne0$). Apply the additive degree formula:

Since $\deg f\ge0$ and $\deg g\ge0$ are non-negative integers summing to $0$, we must have

Thus $f=a_0$ and $g=b_0$ are constant polynomials, i.e. $a_0,b_0\in R$.

Step 3 — The constant is a unit of $R$

From $fg=1$ with $f=a_0,\ g=b_0$ we get $a_0b_0=1$ in $R$. Hence $a_0$ has a multiplicative inverse $b_0$ in $R$, so $a_0$ is a unit of $R$.

Therefore every unit of $R[x]$ lies in $R$ and is a unit of $R$. $\blacksquare$

Answer