← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Field Extensions; Tower Law; Algebraic Closure · Algebra · asked 2× in 14 yrs · Read the full method →

Question

Show that every algebraically closed field is infinite.

Technique

Contrapositive via the “Euclid-style” polynomial $\prod_{a\in F}(X-a)+1$, which evaluates to $1$ at every field element and so has no root — defeating algebraic closure for any finite field.

Solution

We prove the contrapositive: a finite field is not algebraically closed, by exhibiting a polynomial over it with no root. Recall $F$ is algebraically closed if every non-constant polynomial in $F[X]$ has a root in $F$.

Step 1 — Suppose, for contradiction, $F$ is finite and algebraically closed

Let $F$ be a finite field, say $F=\{a_1,a_2,\ldots,a_q\}$ with $q=|F|\ge1$ elements. (Every field has at least the distinct elements $0$ and $1$, so $q\ge2$; the argument below works for any finite $q$.)

Step 2 — Construct a polynomial with no root in $F$

Define

The product $\prod_i(X-a_i)$ has degree $q\ge1$ and leading coefficient $1$, so $p$ is a non-constant polynomial (degree $q\ge1$), with leading term $X^q$.

Step 3 — $p$ has no root in $F$

Evaluate $p$ at any element $a_j\in F$. The factor $(X-a_j)$ appears in the product, so $\prod_i(a_j-a_i)=0$ (the $i=j$ term is $a_j-a_j=0$). Hence

So $p$ has no root in $F$. (Here $1\ne0$ because in a field $1\ne0$.)

Step 4 — Contradiction

$p$ is a non-constant polynomial in $F[X]$ with no root in $F$. This contradicts the assumption that $F$ is algebraically closed (which requires every non-constant polynomial to have a root in $F$). Therefore no finite field is algebraically closed.

Equivalently, by contraposition: if $F$ is algebraically closed, then $F$ is infinite.

Answer