An Irreducible Element that is Not a Prime

[@taxtr4535]

Well, right! This has helped me do my algebra homework, thank you.

[@Misseldine]

You're welcome.

[@aklenaklen]

Well, wrong! The Z[sqrt(-3)] IS A PID. And consequently a UFD. The error lies in the fact that you use 2 twice, but associates are vorbidden. So up to the definition you need TWO DISTINCT IRREDUCIBLES which are not associated. Besides already Gauss proved that the algebraic rings Z[sqrt(-D)] are for D = 1, 2, 3, 7, 11, 19, 43, 67, and 163 PIDs. But if you use instead the ring Z[sqrt(-5)] and the 6=2*3 and 6 = (1+sqrt(-5))(1+sqrt(-5)) it will become true ;)

[@Misseldine]

Anton, thanks for your comment. If there are any errors in the lectures, then I love it when they are pointed out. Unfortunately (or fortunately, depending on how you look at it), this is not an error. Let me explain the confusion here, which is something I often get mixed up with. To the theorem of Gauss that you reference, what he proved is that for those numbers, the ring of integers of the quadratic extension Q(sqrt(-D)) is a PID. The problem here is that the ring of integers for the field Q(sqrt (-3)) is Z[(1+sqrt (-3))/2], not Z[sqrt(-3)]. The former is, in fact, a PID, as Gauss showed, but Gauss's result does not apply to Z[sqrt(-3)] since it is not the ring of integers of an algebraic number field. You can verify that in the ring of integers (1+sqrt(-3))/2 divides both 2 and 1+sqrt (-3), which implies that neither are irreducible in the ring of integers, so the argument made in this lecture does not apply. On the other hand, the argument in this lecture does show that Z[sqrt(-3)] is not a PID since it is not a UFD because in a UFD all irreducibles are primes and we have provided an irreducible which is not prime. The fact that 2 is a repeated factor in 4 is not a concern. We demonstrated that 2 is irreducible in this ring using the multiplicative norm, that is, if 2 were reducible then there would have to be an element with norm 2, but no such element exists. If 2 were prime, since 2 divides 4 it divides one factor in every factorization of 4, including (1+sqrt(-3))(1-sqrt(-3)). But if 2 divides either of these factors, then by norm considerations, they would have to be associates. Considering that the only units are 1 and -1, they are not associates. Simply put, 2 divides 4 but there is a factorization of 4 for which 2 divides neither factor. Hence, 2 is not prime. Neither definition of irreducible nor prime elements has any restrictions about repeated factors. Thus, there is no error here (or at least not with the present objection).

[@aklenaklen]

@@Misseldine Hi Andrew, thanks for you quick answer. I'm definitely not confusing the Eisenstein integers mentioned by you [ (1+sqrt(-3))/2 ] and the ring Z[sqrt(-3)] Actually they are both Euclidian, so PID and so UFD. The Eisenstein integers are one of the possible ways to prove the Last Fermat's theorem for n=3 (The Kummer theory). But you can do it also using the Z[sqrt(-3)] factorizing a^2+3b^2=(a+b*sqrt(-3))(a-b*sqrt(-3)).which is called Euler's proof. And this works precisely because they are both PID so prime and irreducible for both is the same. I don't know which algebra books are valued in you region, I heard you generally appriciate the Dummit & Foote (3rd edition), so this is what this book says about the irreducibles on page 284: "Suppose r is non zero and not unit. Then r is called irreducible if whenever r=ab, at least one of a or b must be a unit". So it's clear that the factorization a^2 is not accepted.

[@Misseldine]

@@aklenaklen Let us first list what we can agree upon. The ring of Eisenstein integers Z[(-1+sqrt(-3))/2], the same ring of integers for Q(sqrt(-3)), is a PID (it is, in fact, a Euclidean domain). Likewise, we are using the same definition of an irreducible element in a ring, that is, a nonzero, non-unit x such that for any factorization x=ab we have that either a or b is a unit. The example in question here is taken from Tom Judson's Abstract Algebra: Theory and Applications (Example 18.10), which is an open source textbook that can be found here: abstract.ups.edu/aata/domains-section-factorization-domains.html (you can also find a link to my lecture notes in the video description). The definition of an irreducible element (as well as a prime element) used in this lecture came directly from this textbook and agrees with the definition found in Dummit and Foote (which I used when in graduate school and am quite familiar with). While the definition of an irreducible element says nothing about the factor a and b other than they are elements of the ring and that one of them is a unit, your inference that an irreducible cannot be a perfect square is correct. If x is an irreducible such that x = a^2 = aa, then a or a is a unit. Thus, a is a unit. As x is a product of two units, it itself is a unit (x^-1 = a^-2) and hence not an irreducible. We both agree on this. The confusion comes about because of the factorization 4 = 2^2 = 2(2). In the example, the claim is that 2, as a member of Z[sqrt(-3)], is an irreducible element. This is determined using the multiplicative norm on Z[sqrt(-3)], namely v(a+b*sqrt(-3)) = a^2+3b^2. Since v(2)=4 and since an element u of Z[sqrt(-3)] is a unit if and only if v(u) = 1, we see that 2 = ab is reducible if and only if v(a) = v(b) = 2. As there is no such element with norm 2, we conclude that 2 is irreducible. Do we both agree with this argument and that 2 is an irreducible in Z[sqrt(-3)]? Next, we want to show that 2 is NOT prime in Z[sqrt(-3)]. Let a= 1+sqrt(-3) and b = 1-sqrt(-3). Note that ab = 4 = 2(2). This second factorization of 4, which involves 2 twice, shows that 2 | 4. This means that 2 | ab. If 2 is prime, then 2 | a or 2 | b. But v(2) = v(a) = v(b) = 4. Hence, if 2 | a, then a = 2u for some unit u, that is, a and 2 are associates in Z[sqrt(-3)]. But 1 and -1 are the only units in Z[sqrt(-3)] (do we agree those are the only units?), the associates of 2 are 2 and -2, a contradiction. So, 2 does not divide a. Likewise, 2 does not divide b, but since 2 | ab we must conclude that 2 is not prime. Do we both agree with the argument and that 2 is not a prime in Z[sqrt(-3)]? If 2 is irreducible and not prime in Z[sqrt(-3)], then Z[sqrt(-3)] is not a UFD (since all irreducibles are primes in a UFD). As all PIDs are UFDs, this also means that Z[sqrt(-3)] is NOT a PID. From your previous comments, it appears the reason that you believe that Z[sqrt(-3)] is a PID is a result exterior to the one presented in this lecture (we will discuss it shortly). As both statements cannot both be true, you have looked for an error in this argument. If I am understanding your previous comments, you do not believe that 2 is irreducible in Z[sqrt(-3)], but I am not sure why you have that conclusion. The fact that 4 = 2(2) has no bearing on the irreducibility of 2. That factorization is sufficient to show that 4 is reducible in Z[sqrt(-3)], but no one is making a contrary claim. We use the fact that 4 = 2(2) twice. Once is in considering norms of element from Z[sqrt(-3)], but the factorization of the norms takes place in Z, so the structure of Z[sqrt(-3)] has no bearing or significance. This gives that 2 is irreducible in Z[sqrt(-3)]. To show that 2 is not prime in Z[sqrt(-3)], we use 4=2(2) again, but this time viewed as a factorization in Z[sqrt(-3)]. This gives an example of an element, 4, which does not have a unique factorization, that is, 2(2) and ab are two distinct factorizations of 4 in Z[sqrt(-3)]. The fact that 4 is a perfect square in Z[sqrt(-3)] is a red herring. It is an insufficient observation with regard to 2's reducibility or primality. There are other references that show that Z[sqrt(-3)] is not a UFD beyond Judson's book and my lecture. Some top Google hits include: math.stackexchange.com/questions/115934/is-mathbb-z-sqrt-3-euclidean-under-some-other-norm, ramanujan.math.trinity.edu/rdaileda/teach/m4363s07/HW8_soln.pdf, www.numerade.com/ask/question/prove-that-zsqrt-3-is-not-a-principal-ideal-domain-3-60243/, quizlet.com/explanations/questions/prove-that-z-sqrt-3-is-not-a-principal-ideal-domain-da7d707c-40cd-47a3-8a21-e668334bb6b5 If you are still not convinced, let me offer a challenge for you to prove your point. If Z[sqrt(-3)] is a PID, as you claim, then the ideal (2, 1+sqrt(-3)) must be principal. Produce the principal generator of this ideal, aka their gcd, and express it as a linear combination of 2 and 1+sqrt(-3). The existence of a gcd would prove that the lecture's argument is flawed. Regrettably I am not an expert on the history of Fermat's Last Theorem to add much comment to your claim. You mentioned a proof due to Euler, which I believe is included here: en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#n_=_3. A perusal of the proof shows that other than a "critical lemma" the proof requires only elementary number theory, that is, the ring structure of Z. Admittedly, this critical lemma appears to be the key ingredient that you were referring to in your previous comment. I am not familiar enough with this critical lemma to have a guess on how the proof goes, but the mountain of evidence shows that Z[sqrt(-3)] is not a UFD, so however the critical lemma is proven, it cannot utilize that Z[sqrt(-3)] is a PID. I would recommend checking again your proof of this critical lemma as a second examination might reveal the confusion here. Thanks again for your comments. This has been a fruitful exercise for me to reexamine the argument used here to check its validity.

[@aklenaklen]

@@Misseldine Hi Andrew! Well you're right. I indeed confused Q[sqrt(-3)] and Z[sqrt(-3)]. While the first is PID the second is not. I'm sorry ;) My strong belief that the Z[sqrt(-3)] was a PID was indeed based on the Euler's proof of the last Fermat theorem for the n=3 case. First he proved the case n=4 where he just slightly improved the Fermat's own proof. The main idea was to factorize p^2-q^2=(p+q)(p-q). Then he used the idea for the case n=3 as follows: x^3+y^3=2p(p^2+3q^2)=z^3 after facrorizing the p^2+3q^2=(p+q*sqrt(-3))(p-q*sqrt(-3)) he basically used the same technique like in case n=4. It is believed that at this moment the algebraic imaginary numbers were born. After that Kummer used the Eisenstein integers to prove for all primes up to p=23 where the Eisenstein (or Kummer) integers become not PID. Basically if we accept that Z[sqrt(-3)] is not PID we have to reject the Eulers proof. Since it was exactly the Fermat's own failure to recognize that not always primes are irreducibles are the same when he claimed that he has a short proof for the whole last theorem. Well, since I never heard that Euler's proof for the case n=3 was wrong, I'm confused. But again thanks a lot for poinint out my mistake!

[@aklenaklen]

@@Misseldine Well I was too hasty to give up. Look at this article: SOME EXAMPLES OF PRINCIPAL IDEAL DOMAIN WHICH ARE NOT EUCLIDEAN AND sOME OTHER COUNTEREXAMPLES www.emis.de/journals/NSJOM/Papers/38_1/NSJOM_38_1_137_154.pdf On page 148 it has following Theorem 3.2. For D = 1, 2, 3, 7, and 11, the ring R of algebraic integers in the field Q[√−D] is a Euclidean domain with the Euclidean norm ϕ, ϕ(α) = αα. What do you say about that??? :)))