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UPSC 2022 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Subrings and ideals · Algebra · asked 4× in 14 yrs · Read the full method →

Question

Let $R$ be the field of real numbers and $S$ the field of all polynomials $f(x)\in R[x]$ such that $f(0)=0=f(1)$. Prove $S$ is an ideal of $R[x]$. Is $R[x]/S$ an integral domain? Justify.

Technique

Verify ideal axioms; identify the principal ideal $S=(x^2-x)$; quotient is not an integral domain because the generator factors; CRT gives explicit structure.

Solution

Note on phrasing. The question calls $S$ a “field” but $S$ is in fact a subset of $R[x]$ (the polynomials vanishing at both $0$ and $1$). We’re asked to prove it’s an ideal.

Step 1 — Prove $S$ is an ideal of $R[x]$

Non-empty: $0\in S$ (zero polynomial vanishes everywhere).

Closed under subtraction: If $f,g\in S$, then $(f-g)(0)=f(0)-g(0)=0-0=0$, $(f-g)(1)=0-0=0$. So $f-g\in S$.

Closed under multiplication by $R[x]$: If $f\in S$ and $h\in R[x]$, then $(hf)(0)=h(0)f(0)=h(0)\cdot 0=0$, $(hf)(1)=h(1)\cdot 0=0$. So $hf\in S$.

All ideal axioms satisfied. $S$ is an ideal of $R[x]$.

Step 2 — Identify $S$ explicitly

$f\in S$ iff $f(0)=0$ and $f(1)=0$, which (since $R$ is a field) means both $x$ and $(x-1)$ divide $f$. Since $x$ and $(x-1)$ are coprime, their product $x(x-1)$ divides $f$.

So $S=(x(x-1))=(x^2-x)$, the principal ideal generated by $x^2-x$.

Step 3 — Is $R[x]/S$ an integral domain?

$R[x]/S=R[x]/(x^2-x)$.

Recall: a quotient ring $R[x]/(p(x))$ is an integral domain iff $p(x)$ is irreducible (since $R[x]$ is a PID over a field).

$p(x)=x^2-x=x(x-1)$ is reducible in $R[x]$ (factors into two linear pieces).

So $R[x]/S$ is not an integral domain.

Explicit zero divisors: Let $\bar x$ and $\overline{x-1}$ be the cosets in $R[x]/S$. Then $\bar x\cdot\overline{x-1}=\overline{x(x-1)}=\overline{x^2-x}=\bar 0$ in $R[x]/S$, but $\bar x\ne\bar 0$ and $\overline{x-1}\ne\bar 0$.

These are non-trivial zero divisors. $R[x]/S$ is not an integral domain.

Answer