# Let $L/K$ be a field extension and let $a, b \in L$ be algebraic elements over $K$ having the same minimal polynomial. Show that $K(a) \simeq K(b)$.

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?

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

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.