# Math Help - Splitting fields

1. ## Splitting fields

I'm trying to prove that splitting fields are unique up to isomorphism.

Say $\Sigma$ is a splitting field for a polynomial $f$ over a field $K$. Suppose also that $\Sigma' \supset K'$ is an isomorphic extension, ie that we have $i:K\longrightarrow K'$ and $j:\Sigma \longrightarrow \Sigma'$ isomorphisms with $j|_{K} = i$.

How can I show that $\Sigma'$ is a splitting field of $i(f)$ over $K'$?

I feel this should be intuitively obvious and that it should fall easily out of definitions and very basic facts, yet I can't seem to be able to prove it nicely.

Thanks.

I'm trying to prove that splitting fields are unique up to isomorphism.

Say $\Sigma$ is a splitting field for (a polynomial?) $f$ over a field $K$. Suppose also that $\Sigma' \supset K$(K'?)is an isomorphic extension, ie that we have (an isomorphism?) $i:K\longrightarrow K'$ and $j:\Sigma \longrightarrow \Sigma'$ isomorphisms with $j|_{K} = i$.

How can I show that $\Sigma'$ is a splitting field of $i(f)$ over $K'$?

I feel this should be intuitively obvious and that it should fall easily out of definitions and very basic facts, yet I can't seem to be able to prove it nicely.

Thanks.
This is the sketch of the proof.

Let $f \in K[x]$ and use induction on $n = [\Sigma: K]$.

For $n=1,\Sigma = K$ and f splits over K, which imples that $i(f)$ splits over $K'$. Thus $\Sigma'=K'$.

If n > 1, assume by induction that $f \in K[x]$ of degree less than n holds the given isomorphism extension property. Then f must have an irreducible factor g of degree greater than 1. Let $\alpha$ be a root of g in $\Sigma$ such that $f=gf_1$, where $f_1$ has degree n-1. Verify that

$K(\alpha) \cong \frac{K[x]}{(g)} \cong \frac{K'[x]}{(g')} \cong K'(\alpha')$, where i sends $\alpha$ to $\alpha'$ and g to g' with coefficients.

Since $\Sigma$ is a splitting field of f over $K(\alpha)$ and $\Sigma'$ is a splitting field of i(f) over $K'(\alpha')$, the induction hypothesis implies that j extends to an isomorphism $\Sigma \cong \Sigma'$.

3. Originally Posted by aliceinwonderland
This is the sketch of the proof.

Let $f \in K[x]$ and use induction on $n = [\Sigma: K]$.

For $n=1,\Sigma = K$ and f splits over K, which imples that $i(f)$ splits over $K'$. Thus $\Sigma'=K'$.

If n > 1, assume by induction that $f \in K[x]$ of degree less than n holds the given isomorphism extension property. Then f must have an irreducible factor g of degree greater than 1. Let $\alpha$ be a root of g in $\Sigma$ such that $f=gf_1$, where $f_1$ has degree n-1. Verify that

$K(\alpha) \cong \frac{K[x]}{(g)} \cong \frac{K'[x]}{(g')} \cong K'(\alpha')$, where i sends $\alpha$ to $\alpha'$ and g to g' with coefficients.

Since $\Sigma$ is a splitting field of f over $K(\alpha)$ and $\Sigma'$ is a splitting field of i(f) over $K'(\alpha')$, the induction hypothesis implies that j extends to an isomorphism $\Sigma \cong \Sigma'$.
Did you perform induction by considering $[K(\alpha):\Sigma] < [K:\Sigma]=n$ and thus the field extensions $K(\alpha)\subset\Sigma$ and $K'(\alpha')\subset\Sigma'$ are isomorphic by induction?

By the way I wasn't specifically asking for a proof of uniqueness of splitting fields, so I guess that caused some confusion. But this is still helpful

I'm already assuming $\Sigma \cong \Sigma'$, and I want to show that $\Sigma'$ must be a splitting field for $i(f)$. (I'm sure its a very trivial result)

4. Actually I have figured out how to show that. Does this sound right?

Let $\alpha\in\Sigma$ be a root of $f$. Then $(t-\alpha)|f \Rightarrow j(t-\alpha) | j(f)=i(f)$. Thus $j(\alpha)$ is a root of $i(f)$. So $i(f)$ splits over $\Sigma'$ and we can show that this is indeed the splitting field as if $L' \subseteq \Sigma'$ is the splitting field of $i(f)$ over $K'$, then $j^{-1}(\L')\subseteq \Sigma$ is a splitting field of $f$ over $K$ and therefore $j^{-1}(L')=\Sigma=j^{-1}(\Sigma')$, so $L'$ must be all of $\Sigma'$.

5. Actually I don't think what I did there is quite right as I didn't specifically show that every root of $i(f)$ is contained in $\Sigma'$. How can i get around this? :/

eg if $f$ has distinct roots $\{\alpha_1,...,\alpha_n\}$ then $\{j(\alpha_1),\cdots,j(\alpha_n)\}$ are distinct and hence are all $n$ roots of $i(f)$, but if they're not distinct this argument breaks down.

I'm already assuming $\Sigma \cong \Sigma'$, and I want to show that $\Sigma'$ must be a splitting field for $i(f)$. (I'm sure its a very trivial result)
Well, I think what I had shown in my previous post is an isomorphism between "splitting" fields using induction, which implies that $\Sigma'$ is a splitting field for $i(f)$.
Zeroes of f in $\bar{K}$ (algebraic closure of K) generate its splitting field $\Sigma$ and the zeroes of i(f) in $\bar{K'}$ generate its splitting field $\Sigma'$ and we see that $\alpha'$ is a root of $i(f)$ in $\bar{K'}$ and $\alpha$ is a root of ${i}^{-1}(if)$ in $\bar{K}$. By hypothesis of field isomorphism, the number of zeroes of f in $\Sigma$ and the number of zeroes of i(f) in $\Sigma'$ should be the same (Think of a vector space isomorphism !).