Galois Theory Overview/Example Part 1: Extension Fields. Overview:

Galois Theory Overview/Example Part 1: Extension Fields I ll start by outlining very generally the way Galois theory works. Then, I will work through an example that will illustrate the Fundamental Theorem of Galois Theory, which brings together group theory and field theory in a really cool way. Overview: The main problem that motivated the development of Galois theory was the (un)solvability of the Quintic. Solvability by radicals: The quadratic formula x = b ± b2 4ac demonstrates that any 2 nd 2a degree polynomial can be solved by radicals in the sense that the roots can be expressed in terms of sums, products, and radicals involving the coefficients. Question: Can every 5 th degree polynomial be solved by radicals? The short answer is no because there are 5 th degree polynomials whose Galois group is isomorphic to S5 which is not a solvable group. This question of course raises other questions: What is a Galois group? What does the solvability of a Galois group have to do with whether you can solve a polynomial by radicals? The way it works (basically) is that given a polynomial, p(x), with coefficients in Q, you can find a bigger field containing Q that contains all the roots of p(x). If the smallest such extension of Q (the splitting field of p(x)) is the right kind of extension (an extension by radicals) then p(x) can be solved by radicals. Now, for every extension field, E, of Q we can consider the group of automorphisms of E that hold the elements of Q fixed. This is the Galois group of E over Q, denoted Gal(E/Q). If p(x) doesn t have any multiple roots and E is the splitting field of p(x), then Gal(E/Q) is called the Galois group of p(x). It turns out that there is a very close relationship between the subgroups of the Galois group of p(x) and the subfields of the splitting field E that contain Q. In fact, the subfield lattice is just the subgroup lattice turned upside- down. The solvability of a group is related to the structure of the subgroup lattice, so it is also related to the structure of the subfield lattice, which in turn is related to whether the splitting field is an extension by radicals. The big conclusion, then is that a polynomial is solvable by radicals if and only if its Galois

group is solvable. The polynomial x 5-6x + 3 has Galois group S5. S5 is not solvable because A5 is simple. The End! Example (54.3): Constructing a Splitting Field Consider the polynomial p(x) = x 4 2 in Q[x]. This polynomial is irreducible over Q. What this means is, it can t be written as a product of polynomials in Q[x] of degree less than 4. This can be proved by Eisenstein s criterion: Theorem 23.15: (Eisenstein Criterion) Let p Z be a prime. Suppose that f (x) = a n x n + + a 0 is in Z[x]. If an is not congruent to 0 (mod p), but a i 0(mod p) for all i<n with a0 not congruent to 0 (mod p 2 ), then f(x) is irreducible over Q. In our example, the leading coefficient is 1, so that is not congruent to 0 mod any p. All the middle coefficients are 0 so they are congruent to 0 mod any p. So, this thing will be irreducible according to Eisenstein as long as we can find a prime p such that p divides the leading coefficient 2, but p 2 does not divide the leading coefficient 2. Well, p = 2, clearly does the trick. So this guy is irreducible. 4 In this case, we can also prove it is irreducible by brute force factoring. Let α = 2 be the real positive fourth root of 2. Then the roots of p(x) are α, - α, iα, and - iα, where i = 1. So p(x) = (x - α)(x (- α))(x - iα)(x (- iα)). Since none of these roots are in Q, we certainly could not factor p(x) over Q into linear factors (or a linear times a cubic). The only chance p(x) has of being reducible over Q is if we can factor it into two quadratics. However, since α 2 = work. 2 is not in Q this will never Because p(x) is irreducible, it ends up being minimal in the sense that any other polynomial in Q[x] that has α as a root will be a multiple of p(x). [This is Theorem 29.13, from the first section on Field Extensions we will go back and hit that section soon.] Now, let s get to work finding an extension field that contains one of the roots of p(x). There are a couple of (equivalent) ways to think about construction such a field. Field Extension Method #1: Toss a root into the field and add enough other stuff to make it a field again. We will start by simply expanding Q to include one extra element, α, to get Q {α}. This new thing is NOT a field. How do we fix that?

Well, first it needs to be closed under multiplication. This means that α 2, and α 3, also need to be included. Note that α 4 = 2 is already included, so all of the other powers of α will be equal to an element of Q times α, α 2, or α 3. All such products will also need to be included in order for our new set to be closed under multiplication. Thus, we need to include any element of the form cα, bα 2, and aα 3 where a, b, and c are elements of Q. For our new set to be a field, it also needs to be closed under addition. So we need to include any possible sum of elements of the Q and the other kinds just listed. This means that we need to include any element of the form aα 3 + bα 2 + cα + d. Elements of this form will add and multiply together much like polynomials using the distributive property. Note that α 4 = 2 will mean that all answers will reduce to a 3 rd degree polynomial. This means that the set will definitely be closed under addition and multiplication now. Is this enough? Because we will multiply these like polynomials, the associative and distributive properties will hold just fine. 0 and 1 are still the additive and multiplicative properties respectively. Addition is clearly still commutative. What about inverses? Note that the inverse of α is!! α3 and the inverse of α 2 is!! α2. So, the guys we chucked in have inverses. It s not so much fun to prove that but I did it with Wolphram Alpha s help. I want to find an inverse for aα 3 + bα 2 + cα + d that looks like wα 3 + xα 2 + cy + z. So, I ll try to solve (aα 3 + bα 2 + cα + d)( wα 3 + xα 2 + cy + z) = 1 for w, x, y, and, z. If you multiply this out and do some substituting (replace α 4 with 2) you will get (dw + cx + by + az)α 3 +(2aw + dx + cy + bz)α 2 +(2bw + 2ax + dy + cz)α +(2cw + 2bx + 2ay + dz) = 1. This gives us a nice system of equations: dw + cx + by + az = 0 2aw + dx + cy + bz = 0 2bw + 2ax + dy + cz = 0 2cw + 2bx + 2ay + dz = 1 We know this thing has a unique solution if the coefficient matrix d c b a 2a d c b 2b 2a d c 2c 2b 2a d is invertible. And Wolfram Alpha says it is (unless a, b, c, and d are zero).

Let s go ahead try to compute an inverse of something interesting but not too complicated, like 2 + 3α. Here d = 2 and c = 3 and the other coefficients are zero. So, we get the matrix: 2 3 0 0 0 2 3 0 0 0 2 3 6 0 0 2 It s obvious by inspection that the inverse of this matrix is: 1 146 8 12 18 27 54 8 12 18 36 54 8 12 24 36 54 8 So we can find our inverse by multiplying this by the vector 0 0 0 1 This gives us the vector 27 146 18 146 12 146 8 146 27 So our inverse is 146 α 3 18 146 α 2 + 12 146 α 8 which is probably what you would have 146 guessed anyways. Let s see if this actually works: (3α + 2) ( 27 146 α 3 18 146 α 2 + 12 146 α 8 146 ) = 81 146 α 4 54 146 α 3 + 36 146 α 2 24 146 α + 54 146 α 3 36 146 α 2 + 24 146 α 16 146 = 162 146 54 146 α 3 + 36 146 α 2 24 146 α + 54 146 α 3 36 146 α 2 + 24 146 α 16 146 = 1 Wow!

The point of all of that was that Wolfram Alpha and I would like you to believe that every non- zero element of the form aα 3 + bα 2 + cα + d has an inverse of this same form. Thus the set of elements of this form is a field! It clearly contains Q as a subfield and it contains 2 roots, α and - α, of p(x). Before, fretting over the other two roots let s talk a bit more about this new field. Notation: This new field is denoted by Q(α). Notice that Q(α) is a vector field over Q because it is closed under addition and scalar multiplication (multiplication by elements of Q). Can we find a nice basis for this vector field? Yes we can! (copy write Barak Obama 2008) Each element is of the form aα 3 + bα 2 + cα + d, so they are all linear combinations of 1, α, α 2, and α 3. So this set, spans Q(α). If this set of vectors was linearly dependent, then there would exist rational numbers a, b, c, d such that aα 3 + bα 2 + cα + d = 0. Well that just means that α is the root of a 3 rd degree polynomial over Q. But any polynomial with α as a root must be divisible by p(x). That means that this must be the zero polynomial (its degree isn t big enough to be a proper multiple). So, one way to think about the Q(α) is as a vector space over Q with basis {1, α, α 2, α 3 }. A word about isomorphisms between field extensions: If I had used the root - α, I would have obtained the exact same field extension. However, if I had use iα (or iα) I would have obtained a different field, Q(iα), consisting of the elements of the form aiα 3 + bα 2 + ciα + d. I claim that the fields Q(α) and Q(iα) are isomorphic. Consider the map φ: Q(α) Q(iα) given by φ(aα 3 + bα 2 + cα + d) = - aiα 3 - bα 2 + ciα + d. This is clearly a bijection because {1, α, α 2, α 3 } and {1, iα, α 2, iα 3 } are bases of the two fields. Now, φ[(aα 3 + bα 2 + cα + d) (wα 3 + xα 2 + yα + z)] = φ[ (dw + cx + by + az)α 3 +(2aw + dx + cy + bz)α 2 +(2bw + 2ax + dy + cz)α +(2cw + 2bx + 2ay + dz)] = - (dw + cx + by + az)iα 3 - (2aw + dx + cy + bz)α 2 +(2bw + 2ax + dy + cz) i α +(2cw + 2bx + 2ay + dz) and φ(aα 3 + bα 2 + cα + d) φ (wα 3 + xα 2 + yα + z) = (- aiα 3 - bα 2 + ciα + d) (- wiα 3 - xα 2 + yiα + z) = - (dw + cx + by + az)iα 3 - (2aw + dx + cy + bz)α 2 +(2bw + 2ax + dy + cz) i α +(2cw + 2bx + 2ay + dz) [Really! I multiplied all of these out and it s really true. You can do it too.]

This observation will be important when we consider our second method of constructing extension fields. This method will also involve constructing a field in which the polynomial x 4 2 has a root, but it won t be the same kind of field. It will actually be made up of cosets. It will however be isomorphic to both Q(α) and Q(iα). The advantage of this first method is that you can construct different but isomorphic field extension something that is needed for Galois theory. Before we learn the second method of constructing field extensions, let s go ahead and finish the job of constructing a splitting field for x 4 2. Q(α) = { aα 3 + bα 2 + cα + d: a, b, c, and d are elements of Q} Notice that every element of Q(α) is a in R, so Q(α) does not contain the roots complex roots iα, and - iα. So, let s extend the field again! Let s extend Q(α) by including i. This will result in a vector space over Q(α) with basis {1, i} (note that i 2 = - 1 so we don t need any higher powers of i to be closed under multiplication). So our new field will consist of the elements of the form a + bi where a and b are elements of Q(α). So, they look like a1α 3 + a2α 2 + a3α + a4 + (a5α 3 + a6α 2 + a7α + a8)i = a1α 3 + a2α 2 + a3α + a4 + a5iα 3 + a6iα 2 + a7iα + a8i. This gives us an 8- dimensional vector field over Q with basis, {1, α, α 2, α 3, i, iα, iα 2, iα 3 }. This new field is denoted Q(α, i) and is a splitting field of p(x) because it contains all of the roots of p(x). Over this field, p(x) can be split (or factored) into linear factors. The degree of an extension E of a field F, denoted by [E : F] is its dimension as a vector field over the base field. [ Q(α, i) : Q] = 8, [ Q(α, i) : Q(α)] = 2, and [ Q(α) : Q] = 4. Radical Extensions: An extension K of a field F is an extension of F by radicals if there are elements α1,, αr K and positive integers n1,, n1, such that K = F(α1,, αr), α1 n1 F, and αi ni F(α1,, αi- 1) for all 1 < i r. A polynomial f(x) F[x] is solvable by radicals over F if the splitting field E of f(x) over F is contained in an extension by radicals. Note that Q(α, i) is a radical extension since α 4 = 2 is in Q and i 2 = - 1 is in Q Q(α).

******************************************************************************************* Time Out: How Do Radical Extensions Relate To Things Like The Quadratic Equation? Recall that the whole point of this exercise is to figure out why there is no analogue to the quadratic equation for 5 th degree polynomials. So, let s try to figure out why radical extensions are connected to solvability by radicals. First, suppose that you have a 2 nd degree irreducible polynomial, f(x), over Q and its splitting field is Q(α) where α 2 is in Q. Let s say c = α 2. Then α = c. Now, we know that all the roots of f(x) are elements of the splitting field Q(α). So they can be expressed in the form a + bα where a, b Q. Substituting, we see that any root is of the form a + b a, b, c Q. This is exactly the form that the roots of a quadratic are expressed in in the quadratic equation. c where For a slightly more complex example, let s consider a 3 rd degree irreducible polynomial, f(x), over Q with splitting field Q(α, β ) where α 3 is in Q, and β 2 is in Q(α). [This is the kind of situation you would expect with an irreducible 3 rd degree polynomial you could extend using a real root first, and then factor out a linear piece and then extend using a root of the 2 nd degree factor.] Anyhow, any root of this polynomial will look like aα 2 + bα + c + dα 2 β + eαβ + fβ, because the basis for Q(α, β ) is {1, α, α 2, β, αβ, α 2 β }. Now, if we let s = α 3, then we can substitute to get any root in the form a( 3 s ) 2 3 3 + b s + c + d( s ) 2 3 β + e s β + fβ. Now if we let lα 2 3 + mα + n = l( s ) 2 3 + m s + n = β 2 3, then β = l( s) 2 3 +m s +n, so, we can express any root in the form: 3 a( s ) 2 3 3 + b s + c + d( s ) 2 3 l( s) 2 3 3 3 +m s +n + e s l( s) 2 3 +m s +n + f 3 l( s) 2 3 +m s +n. As long as the splitting field for the polynomial lives inside a radical extension, we will always be able to do this kind of process. And if it does not live in side a radical extension, we will not be able to do this. *******************************************************************************************

Field Extension Method 2: Form the factor ring Q[x]/<p(x)> Q[x] is the ring of polynomials with coefficients in Q. <p(x)> is the set of all multiples of p(x) in Q[x]. Note that <p(x)> is an ideal. It s a subring because It is clearly non- empty If I subtract one multiple of p(x) from another, I still get a multiple of p(x). And the product of any two multiples of p(x)is clearly a multiple of p(x). It s an ideal because, in fact, the product of any polynomial and a multiple of p(x) is a multiple of p(x). So, Q[x]/<p(x)> is a ring. And because p(x) is irreducible, <p(x)> is maximal and so Q[x]/<p(x)> is a field. [Did you guys do these two theorems in 4/542?] So, what does this thing look like? Well, it consists of cosets, things that look like q(x) + <p(x)>. First, I am going to convince you that any coset can be represented by a polynomial of degree 3 or less. Here goes: We can use the division algorithm to find polynomials d(x) and r(x) where deg r(x) is less than deg p(x) = 4 and q(x) = d(x) p(x) + r(x). This means that q(x) - r(x) is a multiple of p(x) and hence is in <p(x)> so, by the HLCL q(x) + <p(x)> = r(x) + <p(x)>. Now, the opposite is true of polynomials of degree less than 3, those guys will all be in different cosets from each other: Again, by HLCL if q(x) + <p(x)> = q (x) + <p(x)>, then q(x)- q (x) is a multiple of p(x). But if these both have degree less than 4, then this difference will also have degree less than 4, which means that it must be zero. This means that every coset can be associated with one unique representative polynomial that is of degree 3 or less. In other words they all look like (ax 3 + bx 2 + cx + d) + <p(x)>. So, this field kind of looks like our Q(α) from before but in a weird form where everything looks like a polynomial coset thingy. If we take the smallest representative from each coset (the one with degree 3 or less), then drop the <p(x)>, and then change the x s to α s, we get the same kinds of things as were in Q(α). Let s use this idea to prove that p(x) has a root in this new field. I claim that the coset x + <p(x)> is a root. I chose this because if I drop the <p(x)> and replace the x s (there s only one here) with α s, I just get α. So here goes: p(x + <p(x)>) = (x + <p(x)>) 4 2 + = x 4 + <p(x)> 2 = (x 4 2) + <p(x)> = 0. [Note that every time you construct this type of field extension, the coset x + <p(x)> will be a root of p(x) in the new field.]

So this second way also gives us a field that contains Q and a root of x 4 2. [Q[x]/<p(x)> is isomorphic to our first field extension Q(α). All you have to do is map q(x) + <p(x)> to q(α) where q(x) is the smallest (by degree) representative of q(x) + <p(x)>. [This is just mapping x + <p(x)> to α and extending via the homomorphism property.] It s a quick job to prove this is an isomorphism. Now, we can extend this field by forming ([Q[x]/<p(x)>)/< (x 2 +1)+<p(x)>>. This new field will contain a root of x 2 +1. I think it is (x + <p(x)>) + < (x 2 +1)+<p(x)>>. Let s find out! [(x + <p(x)>) + < (x 2 +1)+<p(x)>> ] 2 + [(1+ <p(x)>) + < (x 2 +1)+<p(x)>> ] = [(x + <p(x)>) 2 + < (x 2 +1)+<p(x)>>] + [(1+ <p(x)>) + < (x 2 +1)+<p(x)>> ] = [(x 2 + <p(x)>) + < (x 2 +1)+<p(x)>>] + [(1+ <p(x)>) + < (x 2 +1)+<p(x)>> ] = [ (x 2 + <p(x)>)+ (1+ <p(x)>)] + < (x 2 +1)+<p(x)>> = [(x 2 + 1) + <p(x)>] + < (x 2 +1)+<p(x)>> = 0 + < (x 2 +1)+<p(x)>>. Easy Peasy! This second method gives a second way to think about the degree of a field extension. The degree is the degree of the irreducible polynomial used to construct the field. And if you do it in two steps, you multiply the degrees of the two polynomials you use. So, we again see that our final field ([Q[x]/<p(x)>)/< (x 2 +1)+<p(x)>> is an 8 th degree extension. A Final Note about This Second Method of Constructing a Field Extension Looking back at the two methods we used to construct field extensions, there are a number of differences. One very important difference is that with the first method, we started with a root of our polynomial. Because we were extending Q, we were safe to assume that a root existed in some field (by the Fundamental Theorem of Algebra). In fact, we know enough about C that we were able to factor the polynomial in C and have familiar specific roots to work with. If we didn t know that our field lived inside a field that contains the roots of every polynomial, then we wouldn t have known for sure that there was a larger field that contained a root of our polynomial. The beauty of our second method of construction is that it proves (by construction) the existence of a field containing roots of our polynomial. In fact, this technique works in general to prove that given any polynomial over any field, we can find a larger field that contains a root. This is Theorem 29.3, Kronecker s Theorem.