Prove the equivalence of the following statements regarding field extensions and algebraic elements

Question

I have to prove that following statements are equivalent:

Let K in L be a field extension and a in L. Prove that the following are equivalent:

1. a is algebraic over K
2. the substitution homomorphism $f: K[x]\rightarrow L$ which sends P(x) to P(a) is not injective.
3. $[K[a]:K]$ is finite
4. $\exists m \in \Bbb{N},$ with $a^m\in \langle a ^0,a^1...,a^{m-1}\rangle_K$

So: $1) \rightarrow 2)$ a algebraic, therefore $\exists P\in K[x], P(a)=0$. Therefore the kernel is not 0, and 2) follows.

$2)\rightarrow3)$ i am trying to use the homomorphism theorem. $K[x]$ is a principle ideal doman, therefore the $kerf$ is generated by a single element P(x). Do i then try to prove that there is an isomorphism between $K[x]/\langle P \rangle$ to K[x] and if $P$ is irreducible(why must it be irreducible?) then $[K[x]:K]$ will have the same degree as $P$?

For $3)\rightarrow 4)$ and $4) \rightarrow 1)$ i would really need a hint or help.

Today i am hearing about this for the first time so please be patient with me.

Thanks in advance

Answer 1

$2\to 3$: The image of the map $f:K[x]\to L$ is clearly contained in $K[a]$. Actually, it is equal to $K[a]$. Indeed, every element $k\in K$ is the image of the constant polynomial $p(x)=k$. And the element $a$ is the image of the polynomial $p(x)=x$. So the image (which is obviously a ring) contains the field $K$ and the element $a$, and so it contains $K[a]$. Thus $Im(f)=K[a]$.

Now, let $m(x)$ be the unique monic polynomial which generates the ideal $Ker(f)$. It is clearly irreducible. Indeed, if $m=gh$ where $g,h$ are monic then one of the polynomials $g,h$ belong to $Ker(f)$, and so divisible by $m$. It follows that one of them is equal to $m$, and so $m$ is irreducible. So $K[x]/(m)$ is a field, and by the isomorphism theorems we have $K[x]/(m)\cong K[a]$. Thus $K[a]$ is a field. Moreover, it is easy to check that this is also an isomorphism of vector spaces over $K$. The dimension of $K[x]/(m)$ over $K$ is finite (it is equal to the degree of $m$), and so $[K[a]:K]<\infty$.

$3\to 4$: My hint is that if the dimension of $K[a]$ over $K$ is finite then the set $\{1,a,a^2,a^3,...\}$ is linearly dependent over $K$.

$4\to 1$: Try to write the definition of $a^m$ being a linear combination of $1,a,a^2,...,a^{m-1}$. Once you do this, it should be easy to prove $a$ is the root of some polynomial in $K[x]$.

Answer 2

$\DeclareMathOperator{\im}{im}$(2) $\Rightarrow$ (3):

As you noted, we have $\ker f = (p(x))$ for some nonzero $p(x) \in \ker f$. Note that $K[a]$ is precisely $\im(f)$. We claim that $K[a]$ is generated by $\{1, a, \ldots, a^{n - 1}\}$, where $n := \deg(p(x))$.

This is simple because an arbitrary element of $K[a]$ is of the form $g(a)$ for some $g(x) \in K[x]$. The division algorithm gives $$g(x) = q(x)p(x) + r(x)$$ for polynomials $q(x), r(x) \in K[x]$ with $\deg(r(x)) < \deg(p(x)) = n$. Applying $f$ to the above equation gives $$g(a) = r(a).$$ Since $\deg(r(x)) < n$, we see that $r(a)$ is indeed in the $K$-span of $\{1, \ldots, a^{n - 1}\}$.

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(3) $\Rightarrow$ (4): Since $[K[a] : K]$ is finite, the set $\{1, a, a^2, \ldots\}$ is linearly dependent, but spanning. Conclude.

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(4) $\Rightarrow$ (1): With $m$ as given, note that you can write $$a^m = \alpha_{m - 1}a^{m - 1} + \cdots + \alpha_1 \alpha + \alpha_0$$ for $\alpha_i \in K$. The above shows that $a$ satisfies a polynomial in $K[x]$.

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Although we didn't need this, let us show that $p(x)$ is irreducible, since you asked:

The substitution homomorphism $f : K[x] \to L$ gives you an isomorphism $K[x]/\ker f \cong \im(f)$. (This follows from general ring theory.)

Note that $L$ is an integral domain and thus, so is $\im(f)$. The previous isomorphism then tells us that $p(x)$ is irreducible.