# Fraleigh's proof of Kronecker's Theorem

I was reading Fraleigh's abstract algebra textbook on field extension and he gave a proof of Kronecker's Theorem, but there are several spots that I don't quite understand so any help would be very appreciated.

One question before I start: If $F$ is a field, then is it true that $F$ is a subfield of $F[x]$ since I can view elements in $F$ as constant polynomials in $F[x]?$

Theorem: Let $F$ be a field and $f(x) \in F[x]$ be a nonconstant polynomial. Then there exists an extension field $E$ of $F$ and an $\alpha \in E$ such that $f(\alpha) = 0 .$

Proof: We know that $f(x)$ can be factored into products of irreducible polynomials in $F[x]$ so it is sufficient to prove the theorem when $f(x)$ is irreducible. We define the map $$\psi: F \to F[x]/\langle f(x) \rangle$$ by $\psi(a) = a + \langle f(x) \rangle$ for $a \in F,$ which is a one-to-one map. We defined addition and multiplication in $F[x]/\langle f(x) \rangle$ by choosing any representatives, so we may choose $a \in (a + \langle f(x) \rangle).$ Thus $\psi$ is a homomorphism that maps $F$ one-to-one onto a subfield of $F[x]/\langle f(x) \rangle.$ We identify $F$ with $\{a + \langle f(x) \rangle \mid a \in F \}$ by means of this map $\psi.$ Thus we shall view $E = F[x]/\langle f(x) \rangle$ as an extension field of $F.$

Can someone explain what the bold paragraph is saying? In particular, I don't understand what he meant when he said We defined addition and multiplication in $F[x]/\langle f(x) \rangle$ by choosing any representatives, so we may choose $a \in (a + \langle f(x) \rangle).$ The way I understand it is that since any element $a \in F$ can be written as $a = a + 0.f(x),$ so $F$ is a subfield of $F[x]/\langle f(x) \rangle,$ but I don't think it's the correct interpretation.

Now we show that $E$ contains a zero of $f(x).$ Let $\alpha = x + \langle f(x) \rangle \in E.$ Consider the evaluation homomorphism $\phi_{\alpha}: F[x] \to E,$ then we have: $$\phi_{\alpha}(f(x)) = \phi_{\alpha}(a_{0} + a_{1}x + \dots + a_{n}x^n) = a_{0} + a_{1}(x + \langle f(x) \rangle) + \dots + a_{n}(x + \langle f(x) \rangle)^n \in E$$ where $a_{i} \in F.$

Hence $\mathbf{f(\alpha) = (a_{0} + a_{1}x + \dots + a_{n}x^n) + \langle f(x) \rangle = f(x) + \langle f(x) \rangle = \langle f(x) \rangle = 0 \in E}.$

Regarding the last sentence, $f(x) + \langle f(x) \rangle = \langle f(x) \rangle$ because $f(x) \in \langle f(x) \rangle,$ but why then is $\langle f(x) \rangle = 0?$

Remember how you defined sum and multiplication in factor rings (Section 26). Also as $a=a+0.f(x)$ then $a \in (a+⟨f(x)⟩)$, and you can choose $a$ as representative. For, example $x$ is representative of $x+⟨f(x)⟩$.

So, you are almost right . Since $\psi$ is a monomorphism (why?) , $F$ and $\psi(F)$ are isomorphic and $\psi(F)$ is a subfield of $F[x]/\langle f(x) \rangle$.Then you can see $F$ as a subfield.

In the second part , remember that in $E$: the factor ring, the new zero is $⟨f(x)⟩$ , because in the way you defined the sum in $E$, $⟨f(x)⟩$ is the neutral element.

For your first question, the answer is: yes, it is true that $F$ is a subfield of $F[x]$ since you can view elements in $F$ as constant polynomials in $F[x]$.

Inside $F[x]/\langle f(x)\rangle$, we have for its elements: $$a+\langle f(x)\rangle=a+g(x)f(x)=a$$ for any polynomial $g(x)\in F[x]$. In other words, any polynomial multiple of $f(x)$ is by definition zero inside $F[x]/\langle f(x)\rangle$.

This is equivalent to saying that, inside $F[x]$ i.e. when viewed as polynomials in $F[x]$, $a+g(x)f(x)\equiv a$ modulo $f(x)$, (that is: modulo the equivalence relation on $F[x]$ imposed by the ideal generated by $f(x)$).

So we say that $a$ is a representative of $a+\langle f(x)\rangle$ in $F[x]/\langle f(x)\rangle$. Thus, the author means that the field operations (addition and multiplication) are identified through $a$, which is viewed either as an element of $F$ or as (..a representative of..) an element of $F[x]/\langle f(x)\rangle$. This is why $$\psi: F \rightarrow F[x]/\langle f(x)\rangle \\ a\mapsto \psi(a)=a+\langle f(x)\rangle=a$$ is a monomorphism embedding $F$ inside $E=F[x]/\langle f(x)\rangle$.

Regarding the last sentence, as has already been mentioned above, the ideal $\langle f(x)\rangle$ of $F(x)$ generated by $f(x)$, is identically zero inside $F[x]/\langle f(x)\rangle$.