Useful lemma about monoids and an application

It's a bit annoying to see the same thumbnail over and over on the front page of the blog, so I decided to change it a little bit. My apologies for the small text. I think I used the website hatchful for this one, while the other logo was made with logojoy.

Yesterday, while going through Olympiads, I found two problems that were eerily similar: one from the District Olympiad 2017 and one from the National Olympiad 2011. I shall solve the problem from 2011, while using the problem from 2017 as a lemma.

Let \((A,+,*)\) be a finite ring with the property that \(x^{n+1}+x=0\) has only two roots, 0 and 1. Prove that A is a field.

Solution:

First of all, let us note that, for \(x=1\), \(1+1=0\), so A has characteristic 2. Thus, the order of any element of \(x\) in the additive group \((A,+)\) is either 2 or 1, which implies that the cardinal of A, let's call it n, is a power of 2. Now, let us prove the following lemma:

Lemma:  Let (M,*) be a finite monoid and \(p\) be a natural number bigger than 2, such that \(a^p\) is different from \(a\), for every element \(a\) in M different than the identity element, \(e\). Prove that (M,*) is a group.

Proof: Let us consider such an \(a\). Since M is finite, there will be an \(i\) and a \(k\) , both natural numbers, such that \(a^i=a^{i+k}\). By continuously multiplying with \(k\), we get that \(a^i=a^{i+mk}\), for every \(m\). Thus, for every \(m\), by multiplying with \(a^{mk-i}\), we get that \(a^{mk}=a^{2mk}=a^{pmk}\), so \(a^{pmk}=a^{mk}\), and, thus, \(a^{mk}=e\), so \(a\) is invertible and (M,*) is a group.

Now it's time to apply the lemma to our problem. For any element \(x\) other than 1 and 0, \(x^{n+1}\) is different from \(-x=x\), because the characteristic is 2. By using the proof of the lemma on the monoid (A,*), we get two possibilities: either \(x^i=0\) or \(x^i=1\). All we have to prove now is that 0 is the only nilpotent element of the ring, because all the non-nilpotent elements are invertible.

If \(x\) is a nilpotent element, then let \(i\) be the rank of nilpotence. Thus, \(x^i=0\), while \(x^{i-1}\) is different than 0. By doing calculations, \((x^{i-1}+1)^{n+1}=x^{i-1}+1\) (keep in mind, n is a power of 2, and that is why the expansion is this. If n weren't a power of 2, the result wouldn't hold). Thus, \(x^{i-1}=1\) or \(x^{i-1}=0\), both contradicting the choice of \(x\), which leads to the fact that 0 is the only nilpotent element, which concludes the problem.

The monoid lemma is very useful in abstract algebra. I wonder who proposed the problems, since their similarity is very interesting. You can see, however, that the one from the NMO required a bit more thought, in order to prove that there's only one nilpotent element. Tune in next time, when I will finally post an inequality solved with Lagrange Multipliers. I already have it on paper, but partial derivatives are a bit harder to write than ring theory.

Cris.

Polynomials over a finite field

Oh well, here we go again. Back to Maths posts.

Tragedy struck since the last time we saw, my dear viewers. Oxford rejected me, but there's no time to waste, since the world of Maths is waiting. While going through National Olympiad shortlists, I found the following problem in the shortlist for 2013, randomly sitting at the middle of a page, waiting to be solved. It's a problem about polynomials, so I was surprised when I saw that I could solve it, because I'm not normally great at solving polynomial problems.

Let \(p>1\) be a prime, \((K, +, *)\) be a field with \(p^2\) elements and \(L=[0, 1, ... 1+1+...+1]\), with the last sum having \(p-1\) elements. Prove that for every \(x∊K-L\), there exists \(a,b∊L\) so that \(x^{2013}+ax+b=0\).

Solution:

I'm not sure if there are more ways of solving this particular problem, so for this you really need to have the idea.

Let us presume that there is an element \(x∊K-L\) such that \(x^{2013}+ax+b\) is always different than 0. The first observation is that have \(p^2-p\) elements in \(K-L\). Now, let \(A={(F(X)|F(X)=X^{2013}+aX+b}\). This set has cardinal \(p^2\). Since the cardinal of A is bigger than the cardinal of \(K-L\), we have that at least two polynomials from A have the same value in \(x\).

Let \(f_1\) and \(f_2\) be the associated polynomial functions of the polynomials we found. Thus, \(f_1(x)=f_2(x)\), so \(x^{2013}+a_1 x+ b_1=x^{2013}+a_2 x+ b_2\), so \((a_1 - a_2) x + (b_1 - b_2)=0\). Thus, \(x = (b_2 - b_1)(a_1-a_2)^{-1}∊L\), because \(a_1, a_2, b_1, b_2\) all belong to L and, because p is prime, L is closed under any operation. This contradicts the way we chose x, and thus the problem is solved.

The problem wasn't necessarily hard, but the idea was pretty important. This trick of finding two polynomial functions which have the same value in a point is widely used, though this is the first time I see it used in an abstract algebra problem. Tune in next time, when I might talk about inequalities and analysis, or something else, completely random.

Cris.

Solution to a simplified version of Friedmann's equation - how much has the Universe expanded last year?

Yup, the theory article isn't coming for a while.

Happy New Year, my dear readers! I sincerely hope all your wishes will come true during this wonderful time we're all living in! In the spirit of the New Year, I thought that I might as well calculate how much the Universe has expanded last year. It was either this or a group theory problem with groups of cardinal 2019 taken from Gazeta Matematica. I took this simplified version of the Friedmann equation from Brilliant, which is really great if you want to try hard problems in both Maths and Physics. Without further ado, let's get straight into the problem!

The simplified version of Friedmann's equation states that \((\frac{a'}{a})^2=H_{0}^2(\frac{1-𝛀}{a^3}+𝛀)\), where \(𝛀\) is the proportion of dark energy in the Universe, and \(H_0\) is the Hubble Constant. \(a\) is the size factor, taken such that in 2018 it is 1 (the size factor is the distance between us and a very distant galaxy at a time t over the distance between the same very distant galaxy and us now). In order to find how many times the Universe expanded (at least the known Universe), we must calculate the exact value of the size factor.



First, we are going to make some observations. The size factor is strictly increasing, so \(a'\) can't be 0. Similarly, the size factor is also bigger than 0, because it was 0 only in the initial moment of the Universe.

Back to the problem, let us multiply both sides by \(a^3\), to get rid of the fraction (which makes my Latex job easier). Thus, we get that \((a')^2 a=H_{0}^2 𝛀 a^3+(1-𝛀)H_{0}^2\), and, by moving everything with \(a\) on the left hand side and differentiating with respect to \(a\), we get that
$$2a' a'' a + (a')^3 -3H_{0}^2 𝛀 a' a =0$$
$$⇒ 2 a'' a + (a')^2 - 3H_{0}^2𝛀 a^2=0$$
$$⇒ \frac {2a''}{a}+(\frac{a'}{a})^2=3H_{0}^2𝛀$$

Let us now observe that \((\frac{a'}{a})'=\frac{a''}{a}-(\frac{a'}{a})^2)\), so \(\frac {a''}{a}=(\frac{a'}{a})'+(\frac{a'}{a})^2)\). For the simplification of calculations, let \(y=\frac{a'}{a}\).

$$⇒2y'+3y^2=3H_{0}^2𝛀 ⇒ y'=-1.5y^2+\frac{3H_{0}^2𝛀}{2}$$

As I found out on Wikipedia, this is a simplified version of the Riccati equation (https://en.wikipedia.org/wiki/Riccati_equation), and, as it turns out, its solutions are functions of the form

$$y=-\frac{u'}{3u}, u=A*e^{\frac{3𝛀^{\frac{1}{2}}H_{0}t}{2}}+B*e^{\frac{-3𝛀^{\frac{1}{2}}H_{0}t}{2}}$$

By replacing \(y\) with \(\frac{a'}{a}\) and integrating in t, we get that \(∫\frac{a'}{a}dt=∫-\frac{u'}{3u}dt\).

$$⇒ln(a)+k=-\frac{ln(u)}{3}$$

By taking \(t=t_0\), the current year, we have that $$k=\frac{2 ln(A*e^{\frac{3𝛀^{ \frac{1}{2}}H_{0}t_{0}}{2}}+B*e^{\frac{-3𝛀^{\frac{1}{2}}H_{0}t_{0}}{2}})}{3}$$

By letting t approach 0, we get that \(A=-B\) (this is left as an exercise for the reader)



Thus, after calculations, we reach the final result: $$a(t)=(\frac{e^{\frac{3𝛀^{\frac{1}{2}}H_{0}t}{2}}-e^{-\frac{3𝛀^{\frac{1}{2}}H_{0}t}{2}}}{e^{\frac{3𝛀^{\frac{1}{2}}H_{0}t_{0}}{2}}-e^{-\frac{3𝛀^{\frac{1}{2}}H_{0}t_{0}}{2}}})^{\frac{2}{3}}$$

I promised to find out how much the Universe has expanded during the last year, so here it is. We shall consider that 𝛀=0.68, while \(H_0=\frac{1}{14*10^9}\) years. The Universe's age is about \(13.772*10^9\) years old, and we shall consider that number as \(t_0\), and \(t=t_0+1\). After torturing Wolfram Alpha to tell me the result, here it is; the Universe has expanded 1.0000000000702296877033612965438583887033870149765170 times. I wonder how much that means... how about we find out?

The current size of the Universe is 93 billion light years in diameter. If we consider this to be the size of the Universe in 2018, it means that now, in 2019, the Universe has diameter about 93.000000006531360956412600578578830149414992392816081 billion light years, so it grew by about 000000006531360956412600578578830149414992392816081 billion light years. That's about 6 light years. That might not seem like a lot, but trust me, it is. For example, the distance between the Sun and Alpha Centauri, the closest star, is 4.4 light years. Also, considering that the size factor is a more-or-less the distance between points on a segment, it means that the speed at which the Universe increases is bigger than the speed of light (I won't go into detail about this, because I have no idea how to discuss it). That's a bit too much to comprehend on New Year's.

Happy New Year, folks!

Cris.