SPLITTING FIELDS EITH CONRAD 1. Introuction When is a fiel an f(t ) [T ] is nonconstant, there is a fiel extension / in which f(t ) picks up a root, say α. Then f(t ) = (T α)g(t ) where g(t ) [T ] an eg g = eg f 1. By applying the same process to g(t ) an continuing in this way finitely many times, we reach an extension L/ in which f(t ) splits into linear factors: in L[T ], f(t ) = c(t α 1 ) (T α n ). We call the fiel (α 1,..., α n ) that is generate by the roots of f(t ) over a splitting fiel of f(t ) over. The iea is that in a splitting fiel we can fin a full set of roots of f(t ) an no smaller fiel extension of has that property. Let s look at some examples. Example 1.1. A splitting fiel of T 2 + 1 over R is R(i, i) = R(i) = C. Example 1.2. A splitting fiel of T 2 2 over Q is Q( 2), since we pick up two roots ± 2 in the fiel generate by just one of the roots. A splitting fiel of T 2 2 over R is R since T 2 2 splits into linear factors in R[T ]. Example 1.3. In C[T ], a factorization of T 4 2 is (T 4 2)(T + 4 2)(T i 4 2)(T + i 4 2). A splitting fiel of T 4 2 over Q is Q( 4 2, i 4 2) = Q( 4 2, i). In the secon escription one of the fiel generators is not a root of the original polynomial T 4 2. This is a simpler way of writing the splitting fiel. A splitting fiel of T 4 2 over R is R( 4 2, i 4 2) = R(i) = C. These examples illustrate that, as with irreucibility, the choice of base fiel is an important part of etermining the splitting fiel. Over Q, T 4 2 has a splitting fiel that is an extension of egree 8, while over R the splitting fiel of the same polynomial is an extension (of R!) of egree 2. The splitting fiel of a polynomial is a bigger extension, in general, than the extension generate by a single root. 1 For instance, Q( 4 2, i) is bigger than Q( 4 2). If we are ealing with an irreucible polynomial, ajoining a single root of it to the base fiel always leas to a fiel that, inepenently of the choice of root, is unique up to isomorphism over : if π(t ) is irreucible in [T ] an α is any root of π(t ) in a fiel extension of then = [T ]/(π(t )) by a fiel isomorphism fixing. More precisely, evaluation at α is a surjective homomorphism [T ] fixing an having kernel (π(t )), so there is an inuce fiel isomorphism [T ]/(π(t )). Ajoining a single root of a reucible polynomial to a fiel might lea to non-isomorphic fiels if the root changes. For instance, if f(t ) = (T 2 2)(T 2 3) in Q[T ] then ajoining a single root to Q might result in Q( 2) 1 There is no stanar name in English for the extension of a fiel generate by a single root of a polynomial. One term I have seen use is a root fiel, so Q( 4 2) is a root fiel for T 4 2 over Q. 1
2 EITH CONRAD or Q( 3), an these fiels are not isomorphic. But we shoul expect that any two ways of creating a splitting fiel for (T 2 2)(T 2 3) over Q using all the roots, not just one root shoul lea to isomorphic fiels. Our main task here is to show that something like this is true in general. Theorem 1.4. Let be a fiel an f(t ) be nonconstant in [T ]. If L an L are splitting fiels of f(t ) over then [L : ] = [L : ], there is a fiel isomorphism L L fixing all of (c c for all c ), an the number of such isomorphisms L L is at most [L : ]. Example 1.5. Every splitting fiel of T 4 2 over Q has egree 8 over Q an is isomorphic to Q( 4 2, i). Example 1.6. Every splitting fiel of (T 2 2)(T 2 3) over Q has egree 4 over Q an is isomorphic to Q( 2, 3). This theorem is not only saying that two splitting fiels of f(t ) over are isomorphic, but that they are isomorphic by an isomorphism fixing all of. Our interest in the splitting fiels is not as abstract fiels, but as extensions of, an an isomorphism fixing all of respects this viewpoint. In Theorem 1.4 we are not saying the isomorphism L L fixing is unique. For instance, C an R[x]/(x 2 + 1) are both splitting fiels of T 2 + 1 over R an there are two ifferent isomorphisms R[x]/(x 2 +1) C that fix all real numbers: f(x) mo x 2 +1 f(i) an f(x) mo x 2 + 1 f( i). Our proof of Theorem 1.4 will use an inuctive argument that will only work by proving a stronger theorem, where the single base fiel is replace by two isomorphic base fiels an. Some preliminary ieas neee in the proof are introuce in Section 2 before we get to the proof itself in Section 3. 2. Homomorphisms on Polynomial Coefficients To prove Theorem 1.4 we will use an inuctive argument involving homomorphisms between polynomial rings. Any fiel homomorphism σ : F F extens to a function σ : F [T ] F [T ] as follows: for f(t ) = n i=0 c it i F [T ], set (σf)(t ) = n i=0 σ(c i)t i F [T ]. We call this map applying σ to the coefficients. Writing f(t ) = c n T n +c n 1 T n 1 + + c 1 T + c 0, with c i F, for α F we have σ(f(α)) = σ ( c n α n + c n 1 α n 1 ) + + c 1 α + c 0 (2.1) = σ(c n )σ(α) n + σ(c n 1 )σ(α) n 1 + + σ(c 1 )σ(α) + σ(c 0 ) = (σf)(σ(α)). If f(α) = 0 then (σf)(σ(α)) = σ(f(α)) = σ(0) = 0, so σ sens any root of f(t ) in F to a root of (σf)(t ) in F. Example 2.1. If f(t ) C[T ] an α C then f(α) = f(α), where the overline means complex conjugation an f is the polynomial whose coefficients are complex conjugates of the coefficients of f. If α is a root of f(t ) then α is a root of f(t ).
SPLITTING FIELDS 3 If f(t ) has real coefficients then f(α) = f(α) because f = f when f has real coefficients, an in this case if α is a root of f(t ) then α is also a root of f(t ). Example 2.2. Let F = F = Q(i), σ : F F be complex conjugation, an f(t ) = T 2 + it + 1 3i. In F [T ], (σf)(t ) = T 2 + σ(i)t + σ(1 3i) = T 2 it + 1 + 3i. For any α F, we have σ(f(α)) = σ(α 2 + iα + 1 3i) = σ(α 2 ) + σ(iα) + σ(1 3i) = σ(α) 2 iσ(α) + 1 + 3i, an observe that this final value is (σf)(σ(α)), not (σf)(α). Check that applying σ to coefficients is a ring homomorphism F [T ] F [T ]: σ(f + g) = σf + σg an σ(fg) = (σf)(σg) for any f an g in F [T ], an trivially σ(1) = 1. It also preserves egrees an the property of being monic. If f(t ) splits completely in F [T ] then (σf)(t ) splits completely in F [T ] since linear factors get sent to linear factors. Finally, if σ is a fiel isomorphism from F to F then applying it to coefficients gives us a ring isomorphism F [T ] F [T ], since applying the inverse of σ to coefficients of F [T ] is an inverse of applying σ to coefficients on F [T ]. Example 2.3. Continuing with the notation of the previous example, T 2 + it + 1 3i = (T (1 + i))(t + (1 + 2i)) an σ(t (1 + i))σ(t + (1 + 2i)) = (T 1 + i)(t + 1 2i) = T 2 it + 1 + 3i, which is σ(t 2 + it + 1 3i). 3. Proof of Theorem 1.4 Rather than irectly prove Theorem 1.4, we formulate a more general theorem where the triangle iagram in Theorem 1.4 is expane into a square with isomorphic fiels at the bottom. Theorem 3.1. Let σ : be an isomorphism of fiels, f(t ) [T ], L be a splitting fiel of f(t ) over an L be a splitting fiel of (σf)(t ) over. Then [L : ] = [L : ], σ extens to an isomorphism L L an the number of such extensions is at most [L : ]. σ Theorem 1.4 is a special case of Theorem 3.1 where = an σ is the ientity on. Proof. This proof is long. We argue by inuction on [L : ]. If [L : ] = 1 then f(t ) splits completely in [T ] so (σf)(t ) splits completely in [T ]. Therefore L =, so [L : ] = 1. The only extension of σ to L in this case is σ, so the number of extensions of σ to L is at most 1 = [L : ]. Suppose [L : ] > 1. Since L is generate as a fiel over by the roots of f(t ), f(t ) has a root α L that is not in. Fix this α for the rest of the proof. Let π(t ) be the minimal polynomial of α over, so α is a root of π(t ) an π(t ) f(t ) in [T ]. If there s an isomorphism σ : L L extening σ, then σ(α) is a root of (σπ)(t ): using (2.1), π(α) = 0 σ(π(α)) = σ(0) = 0 ( σπ)( σ(α)) = 0 (σπ)( σ(α)) = 0, where the last step comes from π(t ) having coefficients in (so σ = σ on those coefficients). Therefore values of σ(α) to be etermine must come from roots of (σπ)(t ).
4 EITH CONRAD Now we show (σπ)(t ) has a root in L. Since σ : is an isomorphism, applying σ to coefficients is a ring isomorphism [T ] [T ] (the inverse applies σ 1 to coefficients in [T ]), so π(t ) f(t ) (σπ)(t ) (σf)(t ). Since π(t ) is monic irreucible, (σπ)(t ) is monic irreucible (ring isomorphisms preserve irreucibility). Since (σf)(t ) splits completely in L [T ] by the efinition of L, its factor (σπ)(t ) splits completely in L [T ]. Pick a root α L of (σπ)(t ). Set = eg π(t ) = eg(σπ)(t ), so > 1 (since = [ : ] > 1). This information is in the iagram below, an there are at most choices for α in L. The minimal polynomials of α an α over an (resp.) are π(t ) an (σπ)(t ). (α ) σ There is a unique extension of σ : to a fiel isomorphism (α ) such that α α. First we show uniqueness. If σ : (α ) extens σ an σ (α) = α, then the value of σ is etermine everywhere on because = [α] an ( m ) m m (3.1) σ c i α i = σ (c i )(σ (α)) i = σ(c i )α i. i=0 i=0 In other wors, a -polynomial in α goes to the corresponing -polynomial in α where σ is applie to the coefficients. Thus there s at most one σ extening σ with σ (α) = α. To prove σ exists, we will buil an isomorphism from to (α ) with the esire behavior on an α. Any element of can be written as f(α) where f(t ) [T ] (a polynomial). It can be like this for more than one polynomial: perhaps f(α) = g(α) where g(t ) [T ]. In that case f(t ) g(t ) mo π(t ), so f(t ) = g(t ) + π(t )h(t ). Applying σ to coefficients on both sies, which is a ring homomorphism [T ] [T ], we have (σf)(t ) = (σg)(t ) + (σπ)(t )(σh)(t ), an setting T = α kills off the secon term, leaving us with (σf)(α ) = (σg)(α ). Therefore it is well-efine to set σ : (α ) by f(α) (σf)(α ). This function is σ on an sens α to α. Since applying σ to coefficients is a ring homomorphism [T ] [T ], σ is a fiel homomorphism (α ). For example, if x an y in are written as f(α) an g(α), then xy = f(α)g(α) = (fg)(α) (evaluation at α is multiplicative) so σ (xy) = σ(fg)(α ) = ((σf)(σg))(α ) = (σf)(α )(σg)(α ) = σ (x)σ (y). Using σ 1 : to go the other way shows σ is a fiel isomorphism. Place σ in the fiel iagram below. i=0 σ (α ) σ
SPLITTING FIELDS 5 Now we can finally inuct on egrees of splitting fiels. Take as new base fiels an (α ), which are isomorphic by σ. Since L is a splitting fiel of f(t ) over, it s also a splitting fiel of f(t ) over the larger fiel. Similarly, L is a splitting fiel of (σf)(t ) over an thus also over the larger fiel (α ). Since f(t ) has its coefficients in an σ = σ on, (σ f)(t ) = (σf)(t ). So the top square in the above iagram is like the square in the theorem itself, except the splitting fiel egrees roppe: since > 1, [L : ] [L : ] = < [L : ]. By inuction, [L : ] = [L : (α )] an σ has an extension to a fiel isomorphism L L. Since σ extens σ, σ itself has an extension to an isomorphism L L an [L : ] = [L : ] = [L : (α )] = [L : ]. (If the proof starte with =, it woul usually be false that = (α ), so Theorem 1.4 is not irectly accessible to our inuctive proof.) It remains to show σ has at most [L : ] extensions to an isomorphism L L. First we show every isomorphism σ : L L extening σ is the extension of some intermeiate isomorphism σ of with a subfiel of L. From the start of the proof, σ(α) must be a root of (σπ)(t ). Define α := σ(α). Since σ = σ, the restriction σ is a fiel homomorphism that is σ on an sens α to α, so σ is an isomorphism from to ( σ(α)) = (α ). Thus σ on L is a lift of the intermeiate fiel isomorphism σ := σ. L σ L σ (α ) σ By inuction on egrees of splitting fiels, σ lifts to at most [L : ] isomorphisms L L. Since σ is etermine by σ (α), which is a root of (σπ)(t ), the number of maps σ is at most eg(σπ)(t ) =. The number of isomorphisms L L that lift σ is the number of homomorphisms σ : L lifting σ times the number of extensions of each σ to an isomorphism L L, an that total is at most [L : ] = [L : ]. Example 3.2. Let = = Q an f(t ) = T 4 2, with splitting fiel Q( 4 2, i) over Q. Let α = 4 2, α = i 4 2, σ : be the ientity (the only fiel isomorphism of Q to itself), an σ ( 4 2) = i 4 2. From the proof, there are at most 8/4 = 2 extensions σ of σ to a fiel automorphism of Q( 4 2, i). These are etermine by whether σ(i) can be i or i. Q( 4 2, i) σ Q( 4 2, i) Q( 4 2) σ Q(i 4 2) 4 Q σ 4 Q
6 EITH CONRAD Example 3.3. The fiel Q( 2, 3), a splitting fiel of (T 2 2)(T 2 3) over Q, has egree 4 over Q. In Theorem 3.1 if we use = = Q, L = L = Q( 2, 3), an σ = the ientity map on Q, then there are at most 4 fiel automorphisms of Q( 2, 3) that are the ientity on Q. This oes not guarantee that our boun of 4 is achieve (although it is). 4. The Effect of Separability Now we will a a separability assumption to Theorem 3.1 an get a stronger conclusion: the upper boun [L : ] in Theorem 3.1 for counting isomorphisms L L extening a fixe isomorphism is reache. Theorem 4.1. Let σ : be an isomorphism of fiels, f(t ) [T ], L be a splitting fiel of f(t ) over an L be a splitting fiel of (σf)(t ) over. If f(t ) is separable then there are [L : ] extensions of σ to an isomorphism L L. Proof. The case [L : ] = 1 is easy, so assume [L : ] > 1. As in the previous proof, let α be a root of f(t ) that is not in, π(t ) be its minimal polynomial over, an = eg π(t ). Because f(t ) is separable, (σf)(t ) is separable too. One way to show this is with the characterization of separability in terms of relative primality to the erivative: we can write (4.1) f(t )u(t ) + f (T )v(t ) = 1 for some u(t ) an v(t ) in [T ]. Applying σ to coefficients commutes with forming erivatives (that is, σ(f ) = (σf) ), so if we apply σ to coefficients in (4.1) then we get (σf)(t )(σu)(t ) + (σf) (T )(σv)(t ) = 1, so (σf)(t ) an its erivative are relatively prime in [T ], which proves (σf)(t ) is separable. Any factor of a separable polynomial is separable, so (σπ)(t ) is separable an therefore has roots in L since it splits completely over L. In the proof of Theorem 3.1 we showe the ifferent extensions of σ to a homomorphism σ : L are each etermine by choosing ifferent roots α of (σπ)(t ) in L an letting α α. Since (σπ)(t ) splits completely over L an is separable, it has roots in L, so the number of homomorphisms σ is (an not just at most ). σ (α ) σ Since [L : ] < [L : ] an L is a splitting fiel of f(t ) over, by inuction on the egree of a splitting fiel (along with the new separability hypothesis on f(t )), each σ has [L : ] extensions to an isomorphism L L. Since there are choices for σ, the total number of extensions of σ to an isomorphism L L is [L : ] = [L : ]. Corollary 4.2. If L/ is the splitting fiel of a separable polynomial then there are [L : ] automorphisms of L that fix the elements of. Proof. Apply Theorem 4.1 with =, L = L, an σ the ientity function on.
SPLITTING FIELDS 7 Example 4.3. The extension Q( 2, 3)/Q is a splitting fiel of (T 2 2)(T 2 3). Its egree is 4, so there are 4 automorphisms of Q( 2, 3) fixing Q. This improves Example 3.3 from an upper boun of 4 to an exact count of 4. We will use this a priori count to fin all the automorphisms. Since 2 an 3 generate the extension Q( 2, 3)/Q, any automorphism of the extension is etermine by σ( 2) an σ( 3), which are ± 2 an ± 3. Combining the choices in all possible ways, we get at most 4 choices for σ. See the table below. Corollary 4.2 says there are 4 choices for σ, so every possibility in the table must work. Notice that we raw this conclusion because we know in avance how many σ s exist, before computing the possibilities explicitly. σ( 2) σ( 3) 2 3 2 3 2 3 2 3 Example 4.4. Let = Q(i) an σ be complex conjugation on. The number 1 + 2i is not a square in (check 1 + 2i = (a + bi) 2 has no solution with a, b Q). Thus the fiel L = Q(i, 1 + 2i) has egree 2 over. Complex conjugation on sens T 2 (1 + 2i) to T 2 (1 2i) in [T ], so by Theorem 4.1 with =, complex conjugation on extens in two ways to an isomorphism L L, where L = Q(i, 1 2i). L L 2 σ These two extensions of complex conjugation to an isomorphism L L are etermine by where 1 + 2i is sent: a root of T 2 (1+2i) must go to a root of T 2 (1 2i), so 1 + 2i is sent to a square root of 1 2i in L. Theorem 4.1 tells us that both choices work: each choice arises from an isomorphism L L extening complex conjugation. In Theorem 4.1, it s crucial that L is the splitting fiel of the polynomial (σf)(t ), an not the splitting fiel of f(t ) itself (unless σ is the ientity on ). Theorem 4.1 oes not say that each automorphism σ : extens to an automorphism of a splitting fiel of a polynomial over ; in fact, such an extension to a splitting fiel might not exist if σ is not the ientity on. Example 4.5. Consier the quaratic extension Q( 4 2)/Q( 2). The top fiel is a splitting fiel of T 2 2 over Q( 2). The conjugation automorphism a + b 2 a b 2 on Q( 2) has no extension to an automorphism of Q( 4 2). Inee, we will show every automorphism of Q( 4 2) fixes 2, so it can t restrict to conjugation on Q( 2), which sens 2 to 2. If σ : Q( 4 2) Q( 4 2) is a fiel automorphism, then σ ( 4 2) must be a fourth root of 2 in Q( 4 2). This fiel is insie R, where the only fourth roots of 2 are ± 4 2, so σ ( 4 2) = ± 4 2 for some choice of sign. Regarless of which sign occurs, squaring both sies removes it an leaves us with σ ( 2) = 2 since σ ( 2) = σ ( 4 2 2 ) = (σ ( 4 2)) 2 = (± 4 2) 2 = 2. 2