# Is it true that if $K/F$ is a splitting field and $\phi \colon K \to K'$ is an isomorphism fixing $F$ then $K = K'$?

Let $K$ and $K'$ be finite extensions over a field $F$ such that $K$ is a splitting field for a polynomial $p(x)$ over $F$, and let $\varphi \colon K \to K'$ be an isomorphism which fixes $F$. Is it true that $K = K'$?

This is stated as a theorem in Pinter but I am having some difficulty following the (sparse) proof.

The proof begins with the fact that any such isomorphism takes a root of a polynomial to some other root of the polynomial. So, if $c_1,\ldots,c_n$ are the roots of $p(x)$ in $K$ then $\varphi(c_1),\ldots,\varphi(c_n)$ are the roots of $p(x)$ in $K'$. But Pinter asserts that $\varphi$ permutes the roots, by which I assume he means that $\varphi(c_k) = c_j$ for some $j$ for each $k$. But it's not clear to me why such a statement of equality would even make sense, since $\varphi(c_k) \in K'$ and $c_j \in K$.

I came across this answer by Qiaochu Yuan which seems to hint at where some of my confusion is coming from. It seems like I need to first assume that $K$ and $K'$ are embedded in another field $E$ before even using the symbol "$=$". I would certainly appreciate some clarification on this point. This is why I asked "Is it true?" in the title.

Then, if I understand correctly, we can use the fact that $K$ and $K'$ are simple extensions, say $K = F(a)$ and $K' = F(b)$, and embed them in $F(a,b)$.

I think the next step would be to extend $\varphi$ to an automorphism on $F(a,b)$. Then, since the roots of $p(x)$ are already in $F(a,b)$ and $\varphi$ sends roots to roots, it must permute the roots. But I don't see how to construct such an extension.

Any help would be greatly appreciated.

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Both the answer by Arturo and the answer by Keenan are excellent. I wish I could accept both. – Antonio Vargas Apr 3 '12 at 3:08

It definitely doesn't make sense to assert that $K=K^\prime$ unless you have already embedded them into some other extension of $F$. What you can say is the following: if $L$ is any extension of $F$ at all, and $i_1,i_2:K\rightarrow L$ are two $F$-monomorphisms of $K$ into $L$, then $i_1(K)=i_2(K)$ (the images are the same). This is because $K/F$ is normal. You can assume $p$ is irreducible over $F$. Let $j$ denote either of $i_1,i_2$. If $\alpha$ is a root of $p$ in $K$, then because $j$ is an $F$-monomorphism, $j(\alpha)$ is a root of $p$ in $L$. Let $\alpha_1,\ldots,\alpha_n$ be the distinct roots of $p$ in $K$. Then $K=F(\alpha_1,\ldots,\alpha_n)$, and it follows immediately that $j(K)=F(j(\alpha_1),\ldots,j(\alpha_n))$. Since $p$ splits over $K$, it also splits over $L$, and its distinct roots are the $j(\alpha_k)$. Now observe that none of this depended on $j$. This means that the sets $\{i_1(\alpha_1),\ldots,i_1(\alpha_n)\}$ and $\{i_2(\alpha_1),\ldots,i_2(\alpha_n)\}$ are the same. Thus the images $i_1(K)$ and $i_2(K)$ are the same (they are generated over $K$ by the same set of elements).

If we did not assume that $p$ splits in $K$, then this argument would fail. We know if $p$ splits that it has $n$ distinct roots, and it follows that $p$ has $n$ distinct roots in any extension of $F$ in which it splits. Since the elements $i_1(\alpha_1),\ldots,i_1(\alpha_n)$ are $n$ distinct roots of $p$, they give all the roots of $p$ in $L$. The same argument applies for $i_2$. We have two sets of $n$ distinct roots of $p$, and since $p$ has exactly $n$ distinct roots, these sets must be the same. If $p$ didn't split in $K$, say it only had one root, but it split in $L$ with $3$ distinct roots, then by sending the one root to different roots in $L$, we would get two embeddings with different images.

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Do we have to assume that $j$ fixes $F$ to ensure that $\alpha$ is a root of $p$ $\Longrightarrow$ $j(\alpha)$ is a root of $p$ is true? – Antonio Vargas Apr 3 '12 at 2:54
Yes. Otherwise all you could say is that for a root $\alpha$ of $p$, $j(\alpha)$ is a root of $j(p)$, where $j(p)$ means you apply $j$ to each of the coefficients of $p$ to get a polynomial in $L[X]$. – Keenan Kidwell Apr 3 '12 at 2:59
Thank you Keenan. I really like this approach using inclusions, it makes the argument very clear. – Antonio Vargas Apr 3 '12 at 3:09
You're very welcome! – Keenan Kidwell Apr 3 '12 at 3:12

It is not true that the fields must be "identical".

Consider the two splitting fields of $x^2+1$ over $\mathbb{R}$, one given by the usual $\mathbb{C}$, and another one, $\mathbb{D}$, given by numbers of the form $a+bj$, with $a,b\in\mathbb{R}$ and $j^2=-1$. As sets, these two fields are not equal, though the two fields are certainly isomorphic over $\mathbb{R}$ (by mapping $a+bi\in\mathbb{C}$ to $a+bj\in\mathbb{D}$).

If both $K$ and $K'$ are contained in some field $E$, though, then the argument holds: remember that $p(x)$ has at most $\deg(p)$ roots in any integral domain that contains $F$; in particular, since the roots of $p(x)$ in $E$ are $c_1,\ldots,c_n$, and are also $\varphi(c_1),\ldots,\varphi(c_n)$, then the two sets of roots must be equal; i.e., $\varphi$ is a permutation of the roots that are in $E$.

The next step to remember is that part of the definition of "splitting field" is that $K$ is generated over $F$ by the roots; that is, $K = F(c_1,\ldots,c_n)$, and likewise $K'=F(\varphi(c_1),\ldots,\varphi(c_n))$. But if $\{c_1,\ldots,c_n\}=\{\varphi(c_1),\ldots,\varphi(c_n)\}$ as sets, then clearly we have $$K = F(c_1,\ldots,c_n) = F(\varphi(c_1),\ldots,\varphi(c_n)) = K',$$ giving the desired equality. This is all taking place inside a bigger field $E$.

In this situation, with both $K$ and $K'$ contained in a field $E$, then we do get equality of $K$ and $K'$ as sets, and in particular $\varphi$ is an automorphism of $K$.

In general, though, you cannot do as you propose. Remember that $F(a,b)$ is only defined contextually. In general, given a field $F$, an extension $E$, and a subset $X$ of $E$, $F(X)$ means "the smallest subfield of $E$ that contains both $F$ and $X$." In order to speak about $F(a,b)$, you need to already have an overfield that contains $F$, contains $a$, and contains $b$; you cannot create it ex nihilo simply by saying "consider $F(a,b)$."

(We can get away with it when we adjoin a single element $\alpha$ for the following reason: there is a theorem that says that given a field $F$ and a polynomial $p$, there is an extension of $F$ that contains a root of $p(x)$. So given $\alpha$ that is to be algebraic, if $p(x)$ is the monic irreducible of $\alpha$, we can consider such an extension and let $F(\alpha)$ be the corresponding field there. Then we have another theorem that tells us any two ways of doing this will result in isomorphic, though not necessarily identical, fields. But there is no parallel result for two or more elements)

Again, consider the situation I give above with $F=\mathbb{R}$, $K=\mathbb{R}(i)$, $K'=\mathbb{R}(j)$. There is no field that contains both $K$ and $K'$ as sets; there is no "$\mathbb{R}(i,j)$" to consider. The assertion is false when you do not assume that both splitting fields are contained in the same field; however, you can prove that any two splitting fields of the same (set of) polynomial(s) over $F$ are isomorphic over $F$.

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Thank you Arturo, I was indeed confused on the point about $F(a,b)$. – Antonio Vargas Apr 3 '12 at 3:07
And I am apparently still confused. Why could I not start with $K = F(a)$ and $K' = F(b)$ (as in the question), say that $b$ is algebraic over $F$ so that it is algebraic over $F(a)$, then construct $F(a)(b)$ in the method you mentioned in your parenthetical remark (e.g. by modding $F(a)[x]$ out by the ideal generated by the minimal polynomial of $b$)? Would it not be right to refer to this as $F(a,b)$? – Antonio Vargas Apr 3 '12 at 3:27
@Antonio: Try it with my example again: You first construct $\mathbb{R}[i]$. Then how do you construct $\mathbb{R}[i,j]$? Your original polynomial that defined $j$ need not be irreducible over $\mathbb{R}[i]$, so you cannot guarantee the construction does what you think it does: you don't get a copy of $\mathbb{R}[j]$ inside that extension, because the former assumes that $j$ is a root of an irreducible. If you try doing that construction with my examples, you are forced into $j=i$ or $j=-i$ (depending on which irreducible factor of $x^2+1$ in $\mathbb{R}[i]$ you pick). – Arturo Magidin Apr 3 '12 at 3:37
@Antonio: In other words: you cannot do that construction and guarantee that $a$ an $b$ will remain the distinct objects that you had to begin with. There is a difference between using $a$ and $b$ as placeholder names for "some root of..." and using $a$ and $b$ to represent specific elements of some field. I cannot guarantee the existence of a field $\mathbb{R}(i,j)$ where $i\neq j$, $i\neq -j$ (which I had to begin with) and in which both $i$ and $j$ are roots of $x^2+1$. But I can guarantee the existence of some extension of $\mathbb{R}$ that has roots of $x^2+1$. – Arturo Magidin Apr 3 '12 at 3:43
@Antonio: Again: it depends on whether your $a$ and $b$ are already objects that you have around, or if you are using these names as simple placeholders for roots of polynomials. In the example, $i$ and $j$ are explicit, specific, distinct objects, so we cannot have $j=i$ or $j=-i$; after all, the letter 'j' is not equal to the letter 'i'. These are elements of specific fields $\mathbb{C}$ and $\mathbb{D}$, and I cannot guarantee the existence of a field that contains the field $\mathbb{C}$ as a subset, and the field $\mathbb{D}$ as a subset, with $\mathbb{C}\cap\mathbb{D}=\mathbb{R}$ – Arturo Magidin Apr 3 '12 at 3:49