← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q3a — Step-by-Step Solution

20 marks · Section A

Field Extensions; Tower Law; Algebraic Closure · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Let $K$ be an extension of a field $F$. Prove that the elements of $K$, which are algebraic over $F$, form a subfield of $K$. Further, if $F\subset K\subset L$ are fields, $L$ is algebraic over $K$ and $K$ is algebraic over $F$, then prove that $L$ is algebraic over $F$.

Technique

Reduce “algebraic” to “lies in a finite extension” (Facts B, C), then use the multiplicativity of degree (tower law) to keep extensions finite under field operations and adjunction of finitely many algebraic coefficients.

Solution

The whole proof rests on one lemma: an element $\alpha$ is algebraic over $F$ iff $F(\alpha)$ is a finite extension of $F$, and finite extensions are algebraic, with degrees multiplicative in towers ($[L:F]=[L:K][K:F]$). We use these to close the algebraic elements under field operations and to prove transitivity.

Step 1 — Key facts (finite $\Leftrightarrow$ algebraic-and-finitely-generated; tower law)

Fact A (degree tower law). If $F\subseteq K\subseteq L$ with $[K:F]$ and $[L:K]$ finite, then $L$ is a finite-dimensional $F$-vector space and

Proof sketch. If $\{u_i\}$ is an $F$-basis of $K$ and $\{v_j\}$ a $K$-basis of $L$, then $\{u_iv_j\}$ is an $F$-basis of $L$ (spanning and independence are routine), giving the product of dimensions. $\square$

Fact B. $\alpha\in K$ is algebraic over $F$ $\iff$ $[F(\alpha):F]<\infty$. Proof. If $\alpha$ is a root of an irreducible $p\in F[X]$ of degree $n$, then $F(\alpha)\cong F[X]/(p)$ has $F$-basis $1,\alpha,\ldots,\alpha^{n-1}$, so $[F(\alpha):F]=n<\infty$. Conversely if $[F(\alpha):F]=n<\infty$, the $n+1$ elements $1,\alpha,\ldots,\alpha^{n}$ are $F$-linearly dependent, giving a nonzero polynomial relation, so $\alpha$ is algebraic. $\square$

Fact C. A finite extension is algebraic: if $[E:F]=n<\infty$ and $\beta\in E$, then $1,\beta,\ldots,\beta^n$ are dependent over $F$, so $\beta$ is algebraic over $F$.

Step 2 — Algebraic elements form a subfield of $K$

Let

Clearly $F\subseteq A$ (each $c\in F$ satisfies $X-c=0$), so $A\ne\varnothing$ and $0,1\in A$. Take $\alpha,\beta\in A$; we show $\alpha\pm\beta,\ \alpha\beta,\ \alpha/\beta\ (\beta\ne0)$ are in $A$.

Since $\alpha$ is algebraic over $F$, $[F(\alpha):F]<\infty$ (Fact B). Since $\beta$ is algebraic over $F$, it is a fortiori algebraic over the larger field $F(\alpha)$, so $[F(\alpha,\beta):F(\alpha)]<\infty$. By the tower law (Fact A),

Thus $F(\alpha,\beta)$ is a finite extension of $F$, hence (Fact C) every element of it is algebraic over $F$. In particular

So $A$ is closed under subtraction and under multiplication/division by nonzero elements; together with $0,1\in A$ this makes $A$ a subfield of $K$.

Step 3 — Transitivity: $L/K$ algebraic and $K/F$ algebraic $\Rightarrow$ $L/F$ algebraic

Let $\gamma\in L$ be arbitrary; we must show $\gamma$ is algebraic over $F$.

Since $L$ is algebraic over $K$, $\gamma$ satisfies some polynomial with coefficients in $K$:

Each $k_i\in K$ is algebraic over $F$ (because $K/F$ is algebraic). Build the field

Adjoining finitely many algebraic elements one at a time, each step is a finite extension (Fact B), so by repeated tower law $[F_0:F]<\infty$.

Now $\gamma$ satisfies the above relation with all coefficients in $F_0$, so $\gamma$ is algebraic over $F_0$, giving $[F_0(\gamma):F_0]<\infty$. By the tower law,

Hence $F_0(\gamma)$ is a finite extension of $F$, so $\gamma$ (lying in it) is algebraic over $F$ (Fact C). As $\gamma\in L$ was arbitrary,

Answer