# I need help understanding a proof (Kronecker's theorem)

Kronecker's theorem says that if $$F$$ is a field and $$f(x)$$ is a non-constant polynomial in $$F[x]$$, then there exists an extension field $$E$$ of $$F$$ in which $$f(x)$$ has a root.

Here's the proof provided in the book:

proof:

Since $$F[x]$$ is a UFD, $$f(x)$$ can be expressed as a product of irreducible factors. Consider one of these irreducible factors and call it $$p(x)$$. We need to find an extension in which $$p(x)$$ has a zero. We chose:

$$F[x]/$$

This is a field, since $$p(x)$$ is irreducible.

starting here is where I have questions

Also, we have the inclusion $$F \rightarrow E$$, given by $$a \mapsto a + $$ (I'm not 100% sure how we know this, what allows us to say this map exists?)

Now, $$p(x + ) = \sum_{i = 0}^n a_i(x + )^i = \sum_{i = 0}^n a_ix^i + = p(x) + = 0 + $$

I'm not 100% clear on the last part

If I'm understanding it correctly, this shows that $$x + $$ is a root of p(x), and we can do this for the other irreducible factors as well, but what then? Do we take the union of all of the $$F[x]/$$'s to get $$E$$?

• Any irreducible factor $p(x)$ produces an extension field in which $f(x)$ has a root, which is all we needed to do. Other irreducible factors, if any, are not needed for the proof. Jun 6, 2016 at 19:53
• @Wolverton What "allows" you to say that map exists is that you are showing it and it is well defined, as it is very easy to prove. Jun 6, 2016 at 21:12

Observe that for an element $\;h(x)\in F[x]\;$ ,we have (put $\;I:=\langle p(x)\rangle\;$ for simplicity)

$$f(x)+I=I\iff f(x)\in I\iff f(x)=p(x)k(x)\;,\;\;k(x)\in F[x]\iff p(x)\,\mid\,f(x)$$

and since $\;F\;$ is "naturally" (because "naturally" may depend on the author) embedded in $\;F[x]/I\;$ , we can talk of $\;g(\alpha)\;$ , for $\;g(x)\in F[x]\;,\;\;\alpha\in F[x]/I\;$ , so

$$p\left(x+I\right)=p(x)+I=I\;,\;\;\text{since}\;\;p(x)=1\cdot p(x)\in I$$

and remember that $\;I=\overline 0\;$ is the zero element in the ring (and in this case, field) $\;F[x]/I\;$ .

Your understanding is thus correct, and you need no more to prove Kronecker theorem: you already showed there is a root of $\;p(x)\;$ , which of course is also a root of the original $\;f(x)\;$ .

Let $P_i(X) , 1\leq i\leq m$ the different factors irreducibles of $f$, if $F_i$ is a finite field extension of $F$ s.t $P_i$ has only linear irreducibles factors in $F_i[X]$, then the composite fields $L=F_1F_2\cdot\cdot\cdot F_m$ is a finite extension of $F$ and $f$ has only linear irreducibles factors in $L[X]$. so we must assure the existence of $F_i$.

Let $P=P_i$, the application that you suspect is the composition of the two rings morphisms: injection $F\hookrightarrow F[X]$ and the canonical projection $F[X] \rightarrow F[X]/\langle P(X)\rangle$. this composite morphism is actually injective. as $F [X]$ is principal ($F$ a field ) and $P(X)$ irreducible then $F [X]/\langle P(X)\rangle$ is a field, and the injection $F \rightarrow F [X]/\langle P(X)\rangle$ is a morphism of field ie $F [X]/\langle P(X)\rangle$ is an extension of $F$, and whose $X+\langle P(X)\rangle$ is a root of $P(Y)$. As the degree of $P$ is finite, $P$ admits a finite number of root, so that by repeating a same process a finite number is reached to build an extension of $F$ which $P$ admits all its roots.