From: ah170@FreeNet.Carleton.CA (David Libert) Subject: Re: Are the "experts" lying? re: Simple FLT proof Date: 16 Oct 2000 09:08:40 GMT Newsgroups: sci.math Summary: [missing] "James Harris" (jstev@mindspring.com) writes: > > So far all objections to my proof of FLT center on the assertion (without > proof) that > > x^p + y^p - z^p > > doesn't factor. In any commutative ring define a in the ring is a zero divisor iff a != 0 & there exists b != 0 s.t. a*b = 0 . The polynomial x^p + y^p - z^p in three indeterminates x,y,z makes sense in any ring of polynomials over coefficient ring a commutative ring with unit element. An element of a commutative ring with unit element is called a unit iff it has a multiplicative inverse. An element a of a commutative ring with unit element is defined to be irreducible iff for all ring elements b,c : a = b*c -> b or c is a unit. In the special case of a polynomial ring over coefficients a commutative ring with unit element, a polynomial is a unit iff it is a constant polynomial, with constant coefficient a unit in the coefficient ring. So in such a polynomial ring, a polynomial is irreducible iff any factoring of it has one factor a unit as just described. I will be discussing whether x^p + y^p - z^p as a polynomial in three indeterminates is irreducible over rings of coefficients R. I will be discussing this in some generality, while the cases of most interest to the JSH FLT discussions are subrings of the complexes. I will accordingly be explicit about the assumptions needed on R. So I will first mention examples showing an assumption is necessary. Suppose we have a commutative ring R with unit element 1 s.t. p = 0, ie a sum of p 1's = 0. Then in case p is prime, expanding (x+y-z)^p by the multinomial theorem, all mixed terms get multinomial coefficients divisible by p, and so are zero in R, so in the ring of polynomials in indeterminates x,y,z with coefficients in R we have (x+y-z)^p = x^p + y^p - z^p, so in this case x^p + y^p - z^p is reducible (ie not irreducible). Fermat Claim: for any commutative ring R with unit element 1, with no zero divisors, with p odd and p != 0 in R, in the ring of polynomials in indeterminates x,y,z over coefficients in R, x^p + y^p - z^p is irreducible for p odd. I will prove the Fermat Claim below, after some preliminaries. An element p of a commutative ring with unit element is defined to be prime iff for all a,b in the ring p|a*b -> p|a or p|b . I state and prove Eisenstein's Criterion for irreducibility. This is a method for proving particular polynomials are irreducible in polynomial rings over commutative rings of coefficients. This is discussed in each of the books titled _Algebra_ by respective authors Serge Lang and Thomas Hungerford. My 1974 edition of Hungerford has a typo in the statement of Eisenstein's Criterion. [I noted that last to help avoid confusion for anyone looking that up. But I just realized another benefit: hey Pertti! I found an error in published mathematics!] Lang and Hungerford state Eisenstein over a UFD of coefficients. The proof I will give below is essentially theirs, but from my way of stating it I only seem to need commutative coefficients, as I will state it below. A common factor of the coefficients of a polynomial can always be pulled out as a leading constant polynomial, giving a factoring. So any polynomial with any non-unit common factor can be factored with a constant polynomial this way, refuting irreducibility. A polynomial is called primitive iff all common factos of the coefficients are units in the coefficient ring. All non-primitive polynomials are reducible by this constant factoring, so the criterion will be for primitive polynomials. Eisenstein's Criterion: Suppose S is a commutative ring and r is prime in S. Suppose f = (sum i = 0,n) f_i * X^i is a primitive polynomial in indeterminate X with coefficients in R. Suppose ~r | f_n , for 0 <= i < n r | f_i and ~ r^2 | f_0 Then: f is irreducible in the ring of X indeterminate polynomials over coefficients S Proof of Eisenstein's criterion: Suppose f = g*h, for g and h polynomials in indeterminate X and coefficients in S. Let g_i and h_i enumerate in increasing X powers the respective coefficients of g and h, i running from 0 to the respective degrees of g and h. f_0 = g_0 * h_0, equating constant coefficients. r | f_0, so since r is assumed prime, r | g_0 or r | h_0. Switching names of g and h if necessary, assume without loss of generality r | g_0 . Since we also assumed ~ r^2 | f_0, we also have ~ r | h_0 . We assumed ~ r | f_n , the leading f coefficient. If r divided every g coefficient then r | f_n. So let i be the smallest s.t. ~ r | g_i. Consider the X^i power in g*h. It is a sum including g_i * h_0. i was picked so ~r | g_i, and above we concluded ~ r | h_0. So by the primeness of r, ~ r | (g_i * h_0). All the other summands are of the form g_j * h_(i-j), for 0 <= j (X - s) | f Proof of Root Claim: -> : X - s is a linear monic polynomial, so we can do polynomial division of X - s into f, obtaining a quotient polynomial and constant remainder: f = q*(X - s) + r, for q a polynomial and r in S. Since f(s) = 0, the right side applied to s evaluates to 0. But (X - s) evaluated on s is 0, so q*(X - s) evaluates to 0 on s, so the right side evalute on s is = r. So 0 = r, applying the above equation to s. Since r = 0, from the division equation (X - s)|f <- : assume (X - s)|f . So there is a quotient polynomial q s.t. q*(X - s) = f Evaluating both sides of this equation on s, the linear factor evaluates to 0, so the left side evaluates to 0. The right side evalutes to f(s). So f(s) = 0. QED Root Claim No ZD Claim: For S a commutative ring with no zero divisors S[X] has no zero divisors (S[X] is the ring of polynomials in indeterminate X with coefficients in S) . Proof of No ZD Claim: Suppose f,g non-zero in S[X], that is they are polynomials in indeterminate X with coefficients in S, and neither is the zero polynomial, that is the polynomial with all coefficients zero. The leading coefficient of f*g is the product of the leading coefficients from f and g respectively. By definition of leading coefficient, neither of these is zero. (The hupothesis that neither f nor g were the zero polynomial implies these leading coefficients exist). Since S is assumed to have no zero divisors, this product of two non-zeroes is non-zero. So the leading coefficient of f is non-zero, so f is not the zero polynomial. So non-zero polynomials multiply to a non-zero polynomial, so S[X] has no zero divisiors. QED No ZD Claim Primeness Claim: for R a commutative ring with unit element in R[x,y], x + y is prime Proof of primeness claim: Suppose (x+y) | f*g, for f,g in R[x,y]. I seek to prove (x+y)| f or (x+y)| g . Using the isomorphism R[x,y] to R[x][y], by collecting like powers of y consider these three polynomials as polynomials in indeterminte y, with coefficients in R[x]. Then x+y is a linear polynomial, with linear coefficient 1, and constant coefficient x. This linear polynomial in R[x][y] has root -x, -x as a member of R[x], the ring of coefficients. By the divisibility assumed, -x is a root of f*g, this read as a member of R[x][y]. Evaluating f*g on -x, that is replacing the indeterminate y by the coefficient ring member -x, we get f(-x) * g(-x) where f(-x) and g(-x) are the corresponding evaluations of f and g on -x (replace y by -x). This f(-x)*g(-x) is a product in R[x]. Since -x is a root of f*g by the divisibility, f(-x)*g(-x) = 0 in R[x]. By the No ZD Claim, R[x] has no zero divisors, so one of f(-x) or g(-x) must be 0 in R[x]. Whichever of f or g this is, say f for definite notation, in R[x][y] we therefore have (x+y)|f. This for arbitrary f,g , so x+y is prime. QED Primeness claim For p odd integer, x^p + y^p = (x+y) * (sum 0 <= i < p) x^i * (-y)^(p-i-1) . First Non-divisibility Claim: For R any commutative ring with unit element with no zero divisors and with p odd integer and p != 0 in R, in R[x,y] ~ (x+y) | (sum 0 <= i < p) x^i * (-y)^(p-i-1) Proof of Non-divisibility Claim : Consider these polynomials as in R[x][y]. x+y is a linear polynomial in indeterminate y, so it has root -x. Assume for contradiction (x+y) | (sum 0 <= i < p) x^i * y^(p-i-1) . Then -x would be a root of (sum 0 <= i < p) x^i * (-y)^(p-i-1) . Consider the evaluation of (sum 0 <= i < p) x^i * (-y)^(p-i-1) on -x. Each y is replaced by -x, so this evaluation is (sum 0 <= i < p) x^(p-1) = p*x^(p-1) . We assumed p != 0 in R. So a non-zero coefficient on a pure x power is not the zero polynomial in R[x]. So -x is not a root of (sum 0 <= i < p) x^i * y^(p-i-1) , contradicting the conclusion above. QED First Non-divisibility Claim Second Non-divisibility Claim: for R a commutative ring with unit element, no zero divisors, p an odd integer and p != 0 in R then in R[x,y] ~ (x+y)^2 | x^p + y^p Proof of Second Non-divisibility Claim : Suppose q in R[x,y] and q*(x+y)^2 = x^p + y^p. x^p + y^p = (x+y) * (sum 0 <= i < p) x^i * (-y)^(p-i-1) . So q*(x+y)^2 = (x+y) * (sum 0 <= i < p) x^i * (-y)^(p-i-1) . So (q*(x+y) - (sum 0 <= i < p) x^i * (-y)^(p-i-1) ) * (x+y) = 0 Since R has no zero divisors, by the No ZD Claim, R[x] has no zero divisors. Applying the No ZD Claim again to base ring R[x], and considering R[x][y] as one indeterminate over this, we get R[x][y] has no zero divisors. So R[x,y] has no zero divisors (otherwise collecting like y powers this would induce zero divisors in R[x][y] ). x+y is not the zero polynomial in R[x,y], so the last equation gives the first factor is 0, else we would have zero divisors. So q would also be a quotient of (sum 0 <= i < p) x^i * (-y)^(p-i-1) by x+y, this contradicting the First Non-divisibility Claim. QED Second Non-divisibility Claim Proof of Fermat Claim: Consider x^p + y^p - z^p as a member of R[x,y][z] . Showing irreducibility in R[x,y][z] implies irreducibility in R[x,y,z]. Namely given any factoring in R[x,y,z], rewrite the factors as in R[x,y][z]. As a R[x,y][z] member, x^p + y^p - z^p has leading coefficient -1 and constant coefficient x^p + y^p. x+y divides x^p + y^p in R[x,y], by x^p + y^p = (x+y) * (sum 0 <= i < p) x^i * (-y)^(p-i-1) . By the Primeness Claim, x+y is prime in R[x,y] . By the Second Non-divibility Claim, ~ (x+y)^2 | (x^p + y^p) All coefficients of x^p + y^p - z^p between the leading coefficient (-1) and the constant coefficient (x^p +y^p) are 0, and hence divisible by x+y. ~ (x+y) | -1 in R[x,y], for example viewing x+y in R[x][y], this is a linear polynomial and hence can't divide a constant polynomial. Also, as a R[x,y][z] member, x^p + y^p - z^p is primitive: the leading coefficient is -1 as a R[x,y] member, ie the constant -1 polynomial, which has only unit factors. (Anything dividing -1 must be a unit). So using coefficient ring S = R[x,y], and prime element r = x+y and using x^p + y^p - z^p in R[x,y][z] as a primitive polynomial in one indeterminate over coefficient ring R[x,y], all the premises of the Eisenstein Criterion are met. So x^p + y^p - z^p is irreducible in R[x,y][z], and hence irreducible in R[x,y,z] QED Fermat Claim. We can consider alternate versions of the above. That first version above was considered all variables as indeterminates. Consider now two variables as indeterminates, and one as integer parameter inherited from the start of the proof, or more abstractly for generalization parameter from R. I will discuss the case of parameter x. The other two variants are similar. Earlier with everything as indeterminate I used the same name (x,y,z) for indeterminates as starting variables. Now with variations I will be explicit, and consider parameter x and indeterminates Y and Z. So we are concerned with the polynomial x^p + Y^p - Z^p as a member of R[Y,Z]. If x = 0 this polynomial reduces to a difference of two p-th powers, which can be factored. So assume x != 0. Then the earlier Fermat Claim with the corresponding premises will copy to this new case. Where the previous version worked in R[x,y][z] the new version will work with R[Y][Z]. The previous version in proving properties about R[x,y] processed this as R[x][y] to have the easier case of one indeterminate. In the new version R[Y] is already with one indeterminate and so can be used immediately. Where the previous proof working in R[x][y] took -x for example as in R[x], the new proof will take x in R, x considered as a parameter. One point. The previous proof of the First Non-divisibility Claim had since p is not zero in R, p*x^(p-1) is not the zero polynomial in R[x]. The new proof will correspondingly want to conclude p*x^(p-1) is not zero in R, where x is now a parameter in R. The new proof will still use p not zero in R, and also use the new premise x != 0, and then use no zero divisors in R, to conclude p*x^(p-1) != 0 in R. Finally consider the third case of only one indeterminate, which I will take as Z, the other cases being similar. The polynomial is x^p + y^p - Z^p, where x and y are parameters from R, so x^p + y^p is just a constant. If this constant has a p-th root in R then this polynomial factors by a linear. So for example, if R is the complexes then there is such a p-th root. Well not surprising, one indeterminate polynomials over complexes factor to linear. (Earlier we saw the two and three indeterminate versions don't factor, even when over R = complexes). Or, if R = Z the integers (Z as the set of integers not to be confused with indeterminate Z), then if we are assuming x,y,z are the Fermat counterexample then z is the desired p-th root, and this factors in the one indeterminate version even over integers. On the other hand, if x^p + y^p in integers has any prime factor of exponent 1, then this can be used in Eisenstein, and show x^p + y^p - Z^p doesn't factor over integers. (I was about to write the notation for this polynomial ring: Z[Z] ouch!) So the one indeterminate version goes by cases. The two and three indeterminate cases are uniform: answer no for the premises. Note these premises, p != 0, commutative with unit element, no zero divisors are all very reasonable for the coefficient rings likely to come up in James' proof or variants. -- David Libert (ah170@freenet.carleton.ca) 1. I used to be conceited but now I am perfect. 2. "So self-quoting doesn't seem so bad." -- David Libert 3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig ============================================================================== From: ah170@FreeNet.Carleton.CA (David Libert) Subject: Re: Are the "experts" lying? re: Simple FLT proof Date: 17 Oct 2000 04:50:35 GMT Newsgroups: sci.math Summary: [missing] Last article in this thread I proved that for ring of coefficients R commutative with 1, with p != 0 in R and R with no zero divisors, as a polynomial in indeterminates x,y,z and coefficients in R, x^p + y^p - z^p is indeterminate. (I stated that p was assumed odd, but that was a left over assumption from an earlier draft of the proof. My proof works for p a positive integer.) In the last article I showed p != 0 in R really was necessary, because p = 0 in R gives a non-trivial factoring of x^p + y^p - z^p . I will now give examples showing the no zero divisors premise is necessary, beyond the p != 0 in R premise. For each integer p > 1 I will give an example of a commutative ring R with unit element with p != 0 in R for each integer positive integer p but with zero divisors and with x^p + y^p - z^p reducible in R[x,y,z] for each positive integer p. Namely toward defining R for p>1, first consider Z[a,b,c], the polynomial ring with indeterminates a,b,c and integer coefficients. Take the ideal in Z[a,b,c] generated by the polynomials a^p - 1, b^p - 1, c^p + 1, ab, ac, bc. Let R be Z[a,b,c] mod this ideal. So in R: a^p = 1, b^p = 1, c^p = -1, ab = ac = bc = 0 . There is a normal form for R elements. An element can be represented as an integer + an integer linear combination of each of a,b,c to powers each of 1 through p-1. So an element is determined by 1 + 3(p-1) integers. We define + and * on these normal forms. + adds coefficients termwise. * multiplies out the expressions, taking no crossterms between a,b,c since ab = ac = bc = 0. The powers of each of a,b,c are reduced back to integers or powers of a,b,c in 1,...,p-1 using a^p = 1 etc. We can directly define R as being these normal forms with this definition of +,* , showing that in the original quotient ring definition a,b,c don't collapse together and also don't collapse to any integer. Then in R[x,y,z] : (ax + by + cz)^p = x^p + y^p - z^p . ax + by + cz is not a constant polynomial, so it is not a unit in R[x,y,z], so the factoring above shows x^p + y^p - z^p is reducible. Note that p != 0 in R. -- David Libert (ah170@freenet.carleton.ca) 1. I used to be conceited but now I am perfect. 2. "So self-quoting doesn't seem so bad." -- David Libert 3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig