Proof of Kronecker's Theorem

Kronecker's theorem

Let $F$ be a fiend and $p(x)$ be a non-constant polynomial of $F[x]$. Then there exists an extension field $E$ of $F$ and also there exists some $\alpha \in E$ where $p(\alpha )=0$

Essentially for any polynomial ring of a Field, every polynomial has a zero in an extension field (or $F[x]$)

Proof:

1. First, we only need to check irreducible polynomials as its clear if the polynomial was reducible it would have a zero in the given field.
2. Let $p(x)\in F[x]$ be such a polynomial. Then we have it so $⟨p(x)⟩$ is a maximal ideal in $F[x]$ and thus the quotient ring $F[x]/⟨p(x)⟩$ is a field. We will prove this is the extension field which contains the root of $p(x)$.
3. Now, lets first show that $F[x]/⟨p(x)⟩$ is actually an extension field of $F$ by defining an injective ring homomorphism from $\psi :F\to F[x]/⟨p(x)⟩$ defined as $\psi (f)=f+⟨p(x)⟩$.
4. We can check this is a ring homomorphism as, letting $a,b\in F$$$\psi (a+b)=((a+b)+⟨p(x)⟩)=(a+⟨p(x)⟩)+(b+⟨p(x)⟩)=\psi (a)+\psi (b)$$$$\psi (ab)=(ab+⟨p(x)⟩)=(a+⟨p(x)⟩)(b+⟨p(x)⟩)=\psi (a)\psi (b)$$
5. Lets also prove injectivity:
   1. Assume $\psi (a)=\psi (b)$
   2. then $a+⟨p(x)⟩=b+⟨p(x)⟩$.
   3. Then we have it so $(a-b)+⟨p(x)⟩=0+⟨p(x)⟩$
   4. Notice that this means $a-b\in ⟨p(x)⟩$
   5. The only constant polynomial (and therefor member of $F$) in $⟨p(x)⟩$ is 0 as $⟨p(x)⟩$ is a polynomial with at least deg 1, and as $a,b\in F$ it is clear $a-b=0$ or $a=b$
6. Now lets define F as a subfield of E such that$$\{f+⟨p(x)⟩\mid \mathrm{\forall }f\in F\}$$It is important to understand this holds as every coset only has one polynomial at max, so these are all different cosets.
7. #review