A description of field extensions

Question

Let $k$ be a subfield of $K$ and $a\in K$. The definition of $k(a)$ is $\bigcap\{ \text{subfields of $K$ that contain $k$ and $a$}\}$. Now if $K=k(a)$ then $a$ is called the primitive element. I have seen the following characterization of such subfields as $$ K=k(a) \Leftrightarrow K=\{g(a):g\in k[X]\} $$ now I would like to ask you how to get (formally) to this construct, since I have huge problems understanding it. Somehow it is clear that $k\in\{g(a):g\in k[X]\}$ and because $X\in g[X]$ one can also see how $a\in \{g(a):g\in k[X]\}$ but why is it a (the smallest) field?

Answer 1

Note: The problem is incorrect as stated; you must assume $a$ is algebraic over $k$.

First, note that if $F$ is any subfield of $K$ that contains $a$, and $g(x)\in k[X]$ is a polynomial, then $g(a)\in F$: indeed, if $$g(x) = a_nx^n+\cdots +a_1x+a_0,\qquad a_i\in k$$ then $g(a) = a_na^n+\cdots+a_1a+a_0$ is a sum of products of elements of $F$, and hence must lie in $F$.

Thus, the set $\{g(a)\mid g(X)\in k[X]\}$ is contained in the intersection, and hence contained in $k(a)$.

To prove the other inclusion, we just need to show that the set is a subfield of $K$ that contains $k$ and $a$. You already have verified that it contains $k$ and $a$, so we just need to make sure it is indeed a field. Because if it is a subfield of $K$ that contains $a$ and $k$, then it must contain $k(a)$ by construction.

Most of the properties require no assumptions on $a$; but the existence of multiplicative inverses does.

So, note that the set $F=\{g(a)\mid g(X)\in k[X]\}$ is a ring: it is a subgroup, since we already know it is nonempty and $g(a)-h(a) = (g-h)(a)$, and $(g-h)(X)$ is a polynomial in $k[X]$. And $g(a)h(a) = (gh)(a)$, and $(gh)(x)\in k[X]$, so it is closed under products and thus is a subring of $K$.

The key is multiplicative inverses. This requires $a$ to be algebraic over $k$. Indeed, if $a$ is algebraic over $k$, then let $f(X)$ be the monic irreducible of $a$ over $k$. Note that $g(a)=0$ for $g(X)\in k[X]$ if and only if $f(X)$ divides $g(X)$.

So, let $g(a)\in F$ be nonzero. Then $f(X)$ does not divide $g(X)$, and since $f(X)$ is irreducible in $k[X]$, it follows that $f(X)$ and $g(X)$ are relatively prime. Therefore, there exist polynomial $h(X),m(X)\in k[X]$ such that $1=f(X)h(X) + g(X)m(X)$. Evaluating at $a$, and remembering that $f(a)=0$, we get $1=g(a)m(a)$. But $m(a)\in F$, so $g(a)$ has a multiplicative inverse in $F$, as required.

Thus, $F$ is a subfield of $K$ that contains $k$ and $a$, and therefore $F$ contains $k(a)$. Since we already knew $F$ is contained in $k(a)$, this gives equality.

The statement is false if $a$ is not algebraic, though. If $a$ is transcendental over $k$, then $a$ itself does not have an inverse in $F$. Indeed, suppose there exists $h(X)\in k[X]$ such that $ah(a)=1$. Then the polynomial $Xh(X)-1$ is a polynomial that has $a$ as a root, contradicting the assumption that $a$ is transcendental. Thus, in this case, $F$ is not a field.

When $a$ is transcendental over $k$, the field $k(a)$ is equal to $$\left\{\frac{g(a)}{h(a)}\mid g(X),h(X)\in k[X], h(X)\neq 0\right\},$$ the field of rational functions in $a$.

For general comments on this kind of construction (big intersection vs. explicit description) see this previous answer.