Let $K/F$ be a finite extension. I want to show that $K$ is a splitting field over $F$ $\iff$ any irreducible polynomial $p(x)\in F[x]$ that has a root in $K$ splits completely over $K$.

I haven't even been able to show this in a single direction. My efforts have been to show that if $K$ is the splitting field for some set of polynomials $\{f_1,\ldots,f_r\}\subset F[x]$, than any irreducible polynomial $p(x)\in F[x]$ that has a root in $K$ must split completely. Let $p(\alpha)=0$ with $\alpha\in K$. Then we can factor $p(x)$ over $K$: $p(x)=(x-\alpha)p'(x)$.

From here I am actually stuck. It seems showing $p'(x)$ must have a root in $K$ would be useful but isn't really sufficient to show that it splits completely. My progress on that front is as follows. Letting $\beta_1$ and $\beta_2$ be roots of some irreducible (over $F$) factor of one of the $f_i$, we have an automorphism $\phi$ of $K$ obtained by extending the isomorphism $F(\beta_1)\to F(\beta_2)$ to $K$, which in turn extends the identity map on $F$. This fixes $F$ so fixes $p(x)$. Then we have $\phi((x-\alpha)p'(x))=(x-\phi(\alpha))\phi(p'(x))=p(x)$. Thus, if $\phi(\alpha)\neq\alpha$, then we have a new root for $p(x)$ over $K$. But I don't see why one of these automorphisms necessarily doesn't fix $\alpha$, or how we can show $p(x)$ necessarily splits completely this way.

The book (Dummit and Foote) asks this question before introducing any Galois theory, so I would like to solve it without any of those tools. Thank you.

  • $\begingroup$ What is your definition of a splitting field? $\endgroup$ – k.stm Aug 2 '17 at 19:25
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    $\begingroup$ "The algebraic extension $K/F$ is called a normal extension if it is the splitting field over $F$ for a collection of polynomials in $F[x]$. We shal generally use the term 'splitting field' instead of 'normal extension'" $\endgroup$ – irh Aug 2 '17 at 19:30
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    $\begingroup$ The extension $K/F$ is a splitting field for a polynomial $f(x)\in F[x]$ if $f$ splits completely over $K$ and over no smaller subfield. $\endgroup$ – irh Aug 2 '17 at 19:32
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    $\begingroup$ Is it possible you have already seen a proof that a finite extension $K/F$ is a simple extension $K=F(\alpha)$? $\endgroup$ – hardmath Aug 2 '17 at 19:42
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    $\begingroup$ Can you prove that if $K$ is a splitting field over $F$, letting $F^a$ denote an algebraic closure of $F$, then any $F$-homomorphism $K\to F^a$ is actually an automorphism $K\to K$ ? $\endgroup$ – Maxime Ramzi Aug 2 '17 at 19:55

Not sure if there’s a neat direct way. But I learnt that the trick is to add another more conceptual equivalent statement.

Let $K / F$ be a finite extension of fields and let $L$ be an algebraically closed field containing $K$. So we assume $F ⊆ K ⊆ L$. The following are equivalent:

  1. $K$ is a splitting field for some polynomial $f ∈ F[X]$.
  2. Every field embedding $σ \colon K → L$ fixing $F$ restricts to $K → K$.
  3. Every irreducible polynomial $p ∈ F[X]$ with some root $α ∈ K$ splits completely in $K$.

Proof. (1) ⇒ (2): Let $f ∈ F[X]$ be a polynomial for which $K$ is a splitting field, say of degree $n$ and monic (without loss of generality). Then $f$ has $n$ roots $α_1, …, α_n ∈ K$ (possibly counted with multiplicites) and $f = (X - α_1)·…·(X - α_n)$ in $K$. Then $K = F(α_1, …, α_n)$, as the latter field is a subextension in which $f$ splits and $K$ is by definition the smallest such extension.

Let $σ \colon K → L$ be any field embedding fixing $F$. Then $σ$ map zeroes of $f$ to zeros of $f^σ = f$, that is: $f(\{α_1,…,α_n\}) = \{α_1,…,α_n\}$. As $σ$ fixes $F$ and $K = F(α_1,…,α_n)$, this implies $σ(K) ⊆ K$.

(2) ⇒ (3): Let $p ∈ F[X]$ be any irreducible polynomial with some root $α ∈ K$. We assume it’s nonzero and monic. Then $p$ splits in $L$ because $L$ is algebraically closed, say $p = (X - α_1)·…·(X - α_n)$ for some $α_1, …, α_n ∈ L$. Have a look at $E = F(α)$. Then we have $n$ maps $σ_1, …, σ_n$ all $E → L$ with $σ_i(α) = α_i$ for $i = 1, …, n$. (For this, we need $p$ to be irreducible!) We can extend those to maps $K → L$, which then have to restricct to maps $K → K$ by assumption. As $α_1, …, α_n$ are in the image of $σ_1, …, σ_n$, we therefore have $α_1, …, α_n ∈ K$ and so $p = (X - α_1)·…(X - α_n)$ in $K[X]$.

(3) ⇒ (1): As a finite extension, there are some $β_1, …, β_n ∈ K$ with $K = F(β_1, …, β_n)$ – for example, take an $F$-basis of $K$. Let $p_1, …, p_n ∈ F[X]$ be the minimal polynomials of $β_1, …, β_n$. Being irreducible, all of them split in $K[X]$, so $f = p_1·…·p_n ∈ F[X]$ does a well and $K$ then is the splitting field of $f$.

  • $\begingroup$ Why are we taking $f=p_1\cdots p_n $ ? I think it's correct to say that $K$ is a splitting field of $p_1$ (or any of the $p_i$'s). $\endgroup$ – Shivering Soldier Jan 28 '19 at 17:56
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    $\begingroup$ @ThomasShelby Else, $f$ might split over a smaller subfield $F(β_1,…,β_k) \subsetneq F(β_1, …, β_n)$ for some $k < n$, for instance if $p_1 = (X - β_1)·(X - β_2)$ and $n > 2$.. Hence, that $F(β_1, …, β_k)$, not $F(β_1, …, β_n)$ would be, by definition, the splitting field of $f$. But we need to realise $K = F(β_1, …, β_n)$ itself – nothing less – as a splitting field for some polynomial $f ∈ F[X]$.. $\endgroup$ – k.stm Jan 29 '19 at 13:44
  • $\begingroup$ @k.stm...Why do the maps $\sigma_i$ extend to all of $K$? $\endgroup$ – Indrajit Ghosh Jan 28 '20 at 5:40
  • $\begingroup$ Every field map $σ \colon E → L$ into an algebraically closed field extends to some map $K → L$ for any algebraic extension $K / E$. This is true for simple extensions $K / E$ by the Kronecker construction ($K \cong E[X]/f$ for some irreducible $f ∈ E[X]$) using the universal property of polynomial algebras and can then be extended to any algebraic extensions $K / E$ by some Zorn-style argument, I believe (using pairs $⟨K, σ_K⟩$ of field extensions $K / E$ within an algebraically closed superfield $L$ of $E$ and field map extensions $σ_K \colon K → L$ of $σ$). $\endgroup$ – k.stm Jan 28 '20 at 12:45

I have found a direct way proving this:

Suppose $K$ is the splitting field of $g(x)\in F[x]$ and $p(x)\in F[x]$ is irreducible over $F[x]$. Moreover, $p(x)$ has a root in $K$, we want to show that $p(x)$ splits in $K[x]$.

If $p(x)$ is linear, then we are done.

Otherwise, suppose $p(x)=(x-\alpha)(x-\beta)\tilde p(x)$ in the splitting field $\mathcal F$ of $p(x)$, where $\alpha\in K$, $\tilde p(x)\in\mathcal F[x]$ and $\beta\in\mathcal F\setminus K$ is taken to be any root of $p(x)$ that is not in $K$ and we want to get a contradiction.

Note that there is a natural isomorphism between $F(\alpha)$ and $F(\beta)$, since $p(x)$ is irreducible. Therefore, we can extend this isomorphism naturally to an isomorphism between the splitting field of $g(x)$ over $F(\alpha)$ and the splitting field of $g(x)$ over $F(\beta)$. Since $\alpha\in F$ and we conclude that the splitting field of $g(x)$ over $F(\alpha)$ is $K$. If we let $K'$ denote the splitting field of $g(x)$ over $F(\beta)$, then we have $K\cong K'$ and of course $[K':F]=[K:F]$ for their extension degrees.

Note that $K'$ can be viewed as adjoining $\beta$ to $K$ and $\beta\in\mathcal F\setminus K$ implies that $$[K': F]=[K': K][K: F]>[K:F]$$ which is a contradiction.

Thus, no root of $p(x)$ can be taken from $\mathcal F\setminus K$ which implies $p(x)$ splits completely over $K[x]$.

The other direction is simple, please refer to k.stm's answer for "(3)⇒ (1)".

  • $\begingroup$ Why can $K'$ be viewed as adjoining $\beta$ to $K$? Wouldn't this make $K(\beta)$ bigger than $K$ and therefore not a splitting field for $g(x)$? $\endgroup$ – user46372819 Feb 9 '20 at 2:54
  • $\begingroup$ @AlJebr Because $K'$ being the splitting field of $g(x)$ over $F(\beta)$ means $K'$ is obtained by adjoining all the roots of $g(x)\in F[x]$ over $F$ and the element $\beta$. For your second question, I was arguing by contradiction. $\endgroup$ – Bach Apr 16 '20 at 14:26
  • $\begingroup$ @Bach Am I missing anything or is the sentence starting from "Since $\alpha \in F$..." a misprint? Not sure if it makes sence but I think that $K$ is a splitting for $f$ over $F(\alpha)$ because $K$ contains all of the roots of $f$. $\endgroup$ – Mihail Feb 13 at 3:22

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