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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:

$F[x]/<p(x)>$

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 $a \mapsto a + <p(x)>$ (I'm not 100% sure how we know this, what allows us to say this map exists?)

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

I'm not 100% clear on the last part

The proof ends there with no additional comments.

If I'm understanding it correctly, this shows that $x + <p(x)>$ 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 $F[x]/<p_i(x)>$'s to get $E$?

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  • $\begingroup$ 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. $\endgroup$ Jun 6, 2016 at 19:53
  • $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Jun 6, 2016 at 21:12

2 Answers 2

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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 $\;f(x)\;$ .

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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.

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