The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off

So in a previous question we had to prove that $\mathbb{Q}/(x^{2}-5) \simeq\mathbb{Q}[\sqrt{5}]$, and I did this with the isomorphism $\varphi:\mathbb{Q}/(x^{2}-5)\to\mathbb{Q}[\sqrt{5}],\;ax+b\mapsto a\sqrt{5}+b$

I then thought that I could do the same with this, question, isomorphically mapping $\varphi:K(a) \to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $\psi:K[x]/(f) \to K(b)$ using something of a similar form to my above example. What I came up with was

$$\varphi: K(a) \to K[x]/(f),\; \sum_{i=0}^{n-1}c_{i}a^{i} \mapsto \sum_{i=0}^{n-1}c_{i}x^{i}$$

where $n = \deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) \simeq K[x]/(f) \simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?

  • $\begingroup$ Consider the natural homomorphism from $K [x] $ to $K (a) $ given by $p (x)\mapsto p (a) $. $\endgroup$ – Thomas Shelby Mar 18 at 1:07
  • $\begingroup$ But isn't that not an isomorphism? Does it work both ways? $\endgroup$ – Jack Doherty Mar 18 at 1:18
  • $\begingroup$ The induced map from $K [x]/(f) $ to $K (a) $ will be an isomorphism. Note that $K [x]/(f) $ is a field(why?). $\endgroup$ – Thomas Shelby Mar 18 at 1:20
  • $\begingroup$ It's a field because $f$ is irreducible $\endgroup$ – Jack Doherty Mar 18 at 1:33
  • $\begingroup$ You know that $\phi : K[x] \to K[a]$ being a surjective homomorphism implies $K[x]/\ker(\phi) \cong K[a]$ $\endgroup$ – reuns Mar 18 at 1:44

I agree with the comments. This is just an overview of the formal proof.

Define, $\phi:K[x]\to K(a)$ by $\phi(q(x))=q(a)$ for all $q(x)\in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $\{a\}$. Now it is a homomorphism (check that!) and for any $r\in K(a)$ we set the constant polynomial $q(x)=r,\forall x$ we see $\phi(q(x))=q(a)=r$ so $\phi$ is onto-homomorphism with kernel $Ker(\phi)=\{q(x)\in K[x]:q(a)=0\}=\{q(x):a\text{ is a root of }q(x)\}=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.

Now by, First isomorphism theorem, $$K[x]/(p(x))\cong K(a)$$ Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)\cong K(b))$$ and thus the results follows.

  • 1
    $\begingroup$ Yes, thank you. I had mostly figured it out from the comments. It looks like I was just way overcomplicating it for myself, it's something I tend to have problems with quite a lot. This makes a lot more sense than whatever I was trying to do. $\endgroup$ – Jack Doherty Mar 18 at 13:31

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