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UPSC 2021 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Fields and finite fields · Algebra · asked 6× in 13 yrs · Read the full method →

Question

Let $F$ be a field and $f(x)\in F[x]$ a polynomial of degree $>0$. Show that there exists a field $F'$ and an embedding $q:F\to F'$ s.t. $f^q\in F'[x]$ has a root in $F'$, where $f^q$ replaces each coefficient $a$ by $q(a)$.

Technique

Kronecker’s theorem; construct quotient $F[x]/(f)$ for irreducible $f$; the coset of $x$ is a root.

Solution

Setup. This is the classical Kronecker’s theorem — every non-constant polynomial over a field has a root in some extension field.

Step 1 — Factor $f$ into irreducibles

In $F[x]$, $f=p_1 p_2\cdots p_k$ where each $p_i$ is irreducible in $F[x]$.

It suffices to find an extension where $p_1$ (say) has a root, since then $f$ has a root there too.

So WLOG $f$ is irreducible in $F[x]$.

Step 2 — Construct $F'=F[x]/(f)$

Since $f$ is irreducible in $F[x]$ (a PID), the principal ideal $(f)$ is maximal. So $F'=F[x]/(f)$ is a field.

Step 3 — Embedding $q:F\to F'$

Define $q(a)=a+(f)$ for $a\in F$ (the coset of the constant polynomial $a$).

$q$ is a ring homomorphism: standard.

$q$ is injective: if $q(a)=q(b)$, then $a-b\in(f)$, i.e., $a-b$ is a multiple of $f$. Since $\deg(a-b)=0<\deg f$, the only multiple is $0$, so $a=b$.

So $q$ is an embedding.

Step 4 — $f^q$ has a root in $F'$

The element $\alpha=x+(f)\in F'$ is the coset of $x$.

$f^q(\alpha)=\sum_i q(a_i)\alpha^i=\sum_i(a_i+(f))(x+(f))^i=\sum_i(a_i x^i)+(f)=f(x)+(f)=0+(f)=0_{F'}$.

So $\alpha$ is a root of $f^q$ in $F'$.

Answer