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I've been thinking about my previous question a bit more, and I'm afraid I still don't quite understand. See: Can the natural embedding $K\to K[X]/(f)$ be extended to form an isomorphism $L/K\to K[X]/(f)$?

The theorem I'm trying to prove is as follows: given a field $K$ and an irreducible polynomial $f\in K[X]$, there always exists a (finite) extension $L$ of $K$ such that $L$ contains at least one root of $f$.

Most proofs that I've seen on the internet and in algebra textbooks run like this: $K$ can be embedded in $L'=K[X]/(f)$, which is a field (because $(f)$ is a maximal ideal). So $L'$ is isomorphic to a field extension $L$ of $K$. Then define $\alpha=X\pmod{f}$, and substitute $\alpha$ in $f$. This gives you $0$, so $\alpha$ is a root of $f$.

I get the basic idea, sure enough. But when I try to write a fully rigorous proof, I get stuck.

First of all, literally substituting $\alpha$ in $f$ is impossible, because $\alpha$ is a coset, and not an element of $L$. So, we have to replace the coefficients $a_i$ of $f$ by the corresponding cosets $\overline{a_i}$. This works fine.

But how on Earth do we know for certain that this new polynomial $\sum_i\overline{a_i}X^i$ (which is in $L'[X]$) corresponds, as far as roots are concerned, to the old $f$ (which is in $K[X]$)?

It seems that all proofs I've seen so far implicitly assume that there is an isomorphism $L\to L'$ such that $a_i\mapsto\overline{a_i}$ for all $a_i\in K$. But this is highly non-trivial. Or am I missing something?

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The field generated by the companion matrix of $f$ over $K$ gives you the desired extension.

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  • $\begingroup$ Thanks for offering an alternative line of attack. But the main point is that I want to know why the traditional proofs (with $X\pmod{f}$) work when you look through the `handwaving'. I just don't see it. $\endgroup$ – Kim Fierens Sep 3 '14 at 18:04
  • $\begingroup$ The extension I describe is essentially isomorphic to the $K[X]/(f)$ construct. The isomorphism is given by the map $M^n \leftrightarrow X^n mod (f)$ (with $0\leq i <\deg(f)$) essentially because $f$ is the mininimal polynomial of $M$, or otherwise put $f(M)=0$. "M" is a "root" of $f$, not really a number, but an algebraic entity nevertheless. The embedding of $K$ is done by associating $k \in K$ with the scalar matrix with $k$'s on the diagonal. $\endgroup$ – Marc Bogaerts Sep 3 '14 at 18:28
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The elements of the quotient ring are of the form $p(x)+(f)$ where $p(x) \in K[x]$. And we identify $a$ with $a+(f)$ for $a \in K$ (this isnt a problem is it ?) Now if $q(y) \in K[y] $ and $\overline{q}(y)$ the image in $L[x]$ then $$\overline{q}(p(x)+(f))=q(p(x))+(f)$$

Now this for $q=f$ and $p(x)=x$, this gives

$$\overline{f}(x+(f))=f(x)+(f)=(f)=0$$

So this shows that $x+(f)$ is a root of $\overline{f}$. Under the identification of $K$ with its image $f$ and $\overline{f}$ correspond.

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  • $\begingroup$ My problem is that the identification, let's say $\phi:K\to K[X]/(f)$, of $a$ with $a+(f)$ isn't an isomorphism between the extension $L$ (of $K$) and $K[X]/(f)$. It's only an embedding. So how do we know for sure whether there is an isomorphism $\Psi:K[X]/(f)\to K$ that is compatible with $\phi$? (By compatible I mean: $\Psi^{-1}|_K=\phi$, so that $f(\Psi(\overline{x}))=0\in K$.) $\endgroup$ – Kim Fierens Sep 3 '14 at 19:12
  • $\begingroup$ $\phi$ is an injective homomorphism, so $K[x]/(f)$ is an extension of $\phi(K)$ and $K$ is isomorphic to $\phi(K)$. $\endgroup$ – Rene Schipperus Sep 3 '14 at 19:31
  • $\begingroup$ Yes, up until that point it's perfectly clear to me. But where it gets hazy is this: suppose that $\alpha=\overline{x}$ is our desired root (you and the algebra books say it is). Then there should be an isomorphism $\Psi:K[X]/(f)\to K$ such that $f(\Psi(\alpha))=0$. Right? But at the same time we want that $\Psi(\overline{a_i})X^n+\cdots+\Psi(\overline{a_0})=a_iX^n+\cdots+a_0$. In other words: $\Psi^{-1}|_K =\phi$. It is completely non-obvious to me why such a $\Psi$ should exist. $\endgroup$ – Kim Fierens Sep 3 '14 at 19:46
  • $\begingroup$ There is no such isomorphism $\Phi$. The property that $\alpha$ is a zero is as explained in by answer. $\endgroup$ – Rene Schipperus Sep 3 '14 at 19:51
  • $\begingroup$ I'm very sorry. You've obviously done your best to explain it to me, and I still don't understand. This must be due to some kind of incompetence or blind-spot on my part. I will upvote your answer. $\endgroup$ – Kim Fierens Sep 3 '14 at 19:56

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