ONLY 3 Students Passed?! This Hard Abstract Algebra Exam made 96% of Math Students FAIL!

Today we take a look at yet another university exam where nearly all students failed! This time, it's an abstract algebra and arithmetic exam, that revolves around Group, ring and field theory, matrix diagonalization and elementary number theory. Enjoy! =D

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Postfach 11 15
06731 Bitterfeld-Wolfen
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Germany
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[PapaFlammy]

Make sure to share the video around and to watch the whole thing for the YouTube algorithm to favor this series! :) All the important links can be found at the top of the description btw.! Enjoy! =D

[OmgEinfachNurOmg]

"So why 96% of students failed?"
Papa: "because most students succ balls when it comes to mathematics."

Case closed, don't need to watch the whole video

[maartendj]

I thought by 'algebra' you meant high-school algebra, like solve for X. Watching people fail at that would've been great for my ego. Guess I'll major in English.

[pengfei]

I guess this is a quite reasonable exam assuming students understand all the concepts and study on a regular basis.

[wycked]

Abstract algebra is wicked difficult, only thing I can compare it to in my personal math journey is advanced calculus. Worst part is I chose it as an elective and didn't know what I was getting into.

[thebeerwaisnetwork]

So father got the highest score? Nice.

[tylerrussell]

It's so easy! - Papa
<gets 11 out of 20 points>
B R U H

[KurayamiBotsugo]

"the first part only takes a couple of minutes"

bruh i was figuring out if you should get the determinant or something. Papa flammy roasting us at sucking at math :<

[strtrm]

the meme in the intro is way too easy to read smh

[fabianvonderwarth]

For the eigenvalues..you already showed rank(A+Id) = n/2, so rang Ker(A+Id) =n/2. The Kern is just the Eigenspace. This immediately shows Eigenvalues 1, -1 and that A is diagonizable, since n/2 + n/2 = n. No determinants required

[phoebedraper]

Meme was too easy to pause and read haha. Also I didn't understand much because I'm still in high school, but it was entertaining

[rotiger]

Me: Sees that Papa says 57 is prime Also me: Papa has fallen (I’m writing on mobile so the indentation sucks)

[ElchiKing]

You could have solved ex. 3 even easier: Instead of computing the characteristic polynomial, simply give a basis of Eigenvectors (It's given by etc.) However, I think many people struggled with the topic "Eigenvalue/Eigenvector" (Oh no, that's a difficult topic, I guess the question is hard) or couldn't parse mathematical notation and didn't see it was rather easy
Advice for people watching: If a topic is hard and difficult in general, the question in exams to this topic usually are rather basic. So don't hesitate to do them.


Question 4: This is basically the computation of the radical of an ideal (from a higher point of view).
I guess, the expected answer was something like the following:
1) Prove (maybe by induction in a second semester course) that from p_i|a for every i, it follows p_1*...*p_n|a (1). Then (p_1*...*p_n)^max(k_i) is a multiple of n and a divisor of a^max(k_i). Hence a^(max(k_i))=0 mod n. This is one inclusion
2) If a is in I, then a^k=0 mod n for some k. hence a^k-r*n=0 for some r, or a^k=r*n. Hence, n is a divisor of a^k. In particular, each prime divisor of n is a prime divisor of a^k. Now, we can use Euclids lemma iteratively to deduce p_i|a for every i. This proves the other inclusion.
From (1) we deduce that every Element in I is a multiple of P=p_1*...*p_n. Further p_1*...*p_n is contained in I, so it generates I. The number k is max(k_i) (below, P^k would not be 0, above P^k=0, so every multiple of P will also be 0 when raised to the power k)


It is harder than the others, but actually rather basic definitions (if one likes algebra a bit :P)

[ProfOmarMath]

Recognizing #3 as the matrix representation of the permutation (1 2n)(2 2n-1) .... (n n+1) gives you much of 3 quite directly. For instance, it being a product of disjoint transpositions tells us it has order 2, so its min poly divides x^2-1 and hence splits, so diagonalizable.

[f-th]

Question 4 is about the radical of Z_n (which is I), and it's really easy. If a is divisible by all the prime factors of n, then just raise a to the biggest exponents in the factorization of n. This will make n divide a (because now all the exponents of the primes will be bigger than those of n), so a is in I. For the other direction, take an exponent e that makes a^e = 0. Then that means n divides a^e, so each prime factor of n divides a^e, and so each divides a itself.
Now that you have this result, you can show that the product of the primes is in this ideal, and because Z is a PID, Z_n will be a PID, so it's generated by one element, specifically the smallest one, and clearly the product of the primes is the smallest one (you can't pick any smaller exponents for the generator). Finally, k is the max of the exponents in the factorization of n.

[dradenmerenox]

Last year on my fundamentals of programming midterm, just over half of my class failed. I was in the top 10% with a 70 on it so I was proud of myself for that (It was Waterloo Engineering for reference).

[drumsNgames]

Papa flammy talking about about algebra is great.

[tetraedri_]

I think i have a solution to the last exercise. I will denote elements of Zn in bold.

We are given n=p1^k1•...•pr^kr with pi distinct primes, ki natural numbers. I = { *a* : *a* ^m = 0 for some integer m}.

Claim: *a* is an element of I iff each pi divides a.
Proof: If *a* ^m=0, then each pi^ki|a^m. Also, pi|pi^ki, thus pi|a^m. Each pi is prime, thus pi|a (can be shown by contraposition). This proves one direction
If each pi|a, let k=max(k1, ..., kr). Then pi^k|a^k. Note that k>=ki for each ki, therefore pi^ki|a^k for each ki. Each pi is prime, hence also n|a^k, so *a* ^k=0. This proves the other direction, and finishes the proof.

What is the smallest k such that *a* ^k=0 for each *a* in I?
Solution: in the proof above, we saw that k=max(k1, ...kr) suffices. It is also the smallest: to see this, choose a=p1•...•pr.

Claim: I is a principal ideal domain of Zn generated by *p1•...•pr*
Proof: I is ideal, to see this, raise the sum of two elements of I to a high power and expand. We showed before that for any nonzero *a* in I there exists natural numbers b1, …, br such that a = p1^b1•…•pr^br = We also know that *p1•…•pr* is in I, hence I is generated by *p1•…•pr*

[matron]

Yay! I‘ve been waiting for the next exam Video a while now, and it’s really cool to have early access! Thanks dad!

[DragonKidPlaysMC]

Please do a series on abstract algebra! :))