← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Euclidean domains · Algebra · asked 5× in 13 yrs · Read the full method →

Question

Let $\mathbb K$ be a field and $\mathbb K[X]$ be the ring of polynomials over $\mathbb K$ in a single variable $X$. For a polynomial $f\in\mathbb K[X]$, let $(f)$ denote the ideal in $\mathbb K[X]$ generated by $f$. Show that $(f)$ is a maximal ideal in $\mathbb K[X]$ if and only if $f$ is an irreducible polynomial over $\mathbb K$.

Technique

Use that $\mathbb K[X]$ is a PID (from the division algorithm). Forward: maximal $\Rightarrow$ a nontrivial factorization would create a strictly intermediate principal ideal. Backward: irreducible $\Rightarrow$ any ideal $(g)\supseteq(f)$ collapses to $(f)$ or the whole ring.

Solution

The engine is that $\mathbb K[X]$ is a principal ideal domain (PID): $\mathbb K$ a field $\Rightarrow$ $\mathbb K[X]$ has a division algorithm $\Rightarrow$ every ideal is principal. We must read “maximal” correctly: for $(f)$ to be a proper ideal we need $f\notin\mathbb K^\times\cup\{0\}$, i.e. $f$ non-zero and non-unit, which is also exactly what “irreducible” presupposes.

Step 1 — $\mathbb K[X]$ is a PID; preliminaries

For $g,h\in\mathbb K[X]$ with $h\ne0$, the division algorithm gives $g=qh+r$, $\deg r<\deg h$. A standard consequence: every ideal $I\subseteq\mathbb K[X]$ is principal — if $I\ne(0)$, pick $h\in I$ of least degree; for any $g\in I$, $g=qh+r$ forces $r=g-qh\in I$ with $\deg r<\deg h$, so $r=0$ and $I=(h)$.

Note $(f)=\mathbb K[X]$ iff $f$ is a non-zero constant (a unit), and $(f)=(0)$ iff $f=0$. So $(f)$ is a proper, non-zero ideal exactly when $f$ is a non-zero non-unit, the regime in which irreducibility is defined. Recall: $f$ is irreducible if $f$ is a non-zero non-unit and $f=gh\Rightarrow g$ or $h$ is a unit.

Step 2 — ($\Leftarrow$) $f$ irreducible $\Rightarrow$ $(f)$ maximal

Let $f$ be irreducible. Then $f$ is a non-zero non-unit, so $(f)$ is a proper ideal. Suppose $J$ is an ideal with

Since $\mathbb K[X]$ is a PID, $J=(g)$ for some $g$. From $f\in(g)$ we get $f=gh$ for some $h\in\mathbb K[X]$. By irreducibility, either $g$ is a unit or $h$ is a unit.

• If $g$ is a unit, then $(g)=\mathbb K[X]$, so $J=\mathbb K[X]$.
• If $h$ is a unit, then $g=h^{-1}f\in(f)$, so $(g)\subseteq(f)$, giving $J=(f)$.

Thus the only ideals between $(f)$ and $\mathbb K[X]$ are $(f)$ and $\mathbb K[X]$ themselves: $(f)$ is maximal.

Step 3 — ($\Rightarrow$) $(f)$ maximal $\Rightarrow$ $f$ irreducible

Let $(f)$ be maximal. By definition a maximal ideal is proper, so $(f)\ne\mathbb K[X]$ and $(f)\ne\mathbb K[X]$ also forces $f$ to be a non-unit; and $(f)$ being maximal it is in particular $\ne(0)$ (the whole ring strictly contains $(0)$ and also $(f)\subsetneq\mathbb K[X]$, and $(0)$ is not maximal in $\mathbb K[X]$ since $(0)\subsetneq(X)\subsetneq\mathbb K[X]$). Hence $f$ is a non-zero non-unit.

Suppose, for contradiction, $f=gh$ with neither $g$ nor $h$ a unit. Then $\deg g,\deg h\ge1$ and both are $<\deg f$. Now $f\in(g)$, so

We claim both inclusions are strict.

• $(g)\ne\mathbb K[X]$: else $g$ is a unit, contradiction.
• $(f)\ne(g)$: if $(f)=(g)$ then $g\in(f)$, i.e. $g=fk$ for some $k$, so $f=gh=fkh$, giving $kh=1$ (cancel $f\ne0$ in the domain $\mathbb K[X]$), so $h$ is a unit, contradiction.

Thus $(f)\subsetneq(g)\subsetneq\mathbb K[X]$, contradicting maximality of $(f)$. Hence no such factorization exists: $f$ is irreducible.

Answer