55,199 reputation
470156
bio website users.utu.fi/lahtonen/…
location Rusko, Finland
age 51
visits member for 3 years, 6 months
seen 14 mins ago

General non-sense:

Mostly I teach here. I want to encourage beginning students to think for themselves, so I use a lot of hints and comments. More advanced questions I often just answer.

However, I don't do 1-on-1 chat. Don't ask. My answers and comments are for all to see, so I don't like to hide them. Also I feel that entering a chatroom carries with it an obligation to continue an on-line discussion, and then I would not be using my time on my own terms.

Relevant personal history:

PhD from Notre Dame in '90.

Drifted from representation theory of algebraic groups to applications of algebra into telecommunications, mostly coding theory, and lately mostly teaching at college level.

3 graduate students with awarded PhDs

I have mostly worked at our local University at Turku, Finland. At one point I tried working for Nokia Research Center. It was ok, but an old dog didn't learn all the tricks, and then they downsized, so I returned to the Uni as a tenured lecturer.


42m
comment Question on Galois
Hint: What are the other roots of $x^4-2$? They should all belong to field ____ (you fill in the blank).
16h
comment Prove the automorphism given by $\phi \left(g\right)=\left(g^{-1}\right)^t$ is not an inner automorphism of $SL_n\left(R\right)$
The difference with $n=2$ and $n>2$ can be explained also in terms of Dynkin diagrams. The automorphism $\phi$ is known to correspond to the non-trivial graph automorphism of diagrams of type $A_{n-1}$. This is the identity graph automorphism on the one-point diagram $A_1$ corresponding to $SL_2$. IIRC in general the automorphisms of Chevalley groups and algebraic groups are combinations of an inner automorphism, a field automorphism and a graph automorphism. My upvote was there as soon as I saw your answer.
16h
revised Finding the number that gives remainder equal to 0
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16h
revised Two roots of the polynomial $x^4+x^3-7x^2-x+6$ are $2$ and $-3$. Find two other roots.
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16h
comment Prove the automorphism given by $\phi \left(g\right)=\left(g^{-1}\right)^t$ is not an inner automorphism of $SL_n\left(R\right)$
You need to assume that $n>2$. In $SL_2(\Bbb{R})$ $\phi$ is gotten by conjugation with $$\pmatrix{0&1\cr-1&0\cr}.$$
16h
comment Galois group of $f = x^5 - 2x^3 - x^2 + 2$
The splitting field also has $\Bbb{Q}(\omega)$ as a subfield. As this subfield contains non-real numbers it cannot be equal to $\Bbb{Q}(\sqrt2)$. So the splitting field must have degree $>2$. Also the Galois group must have at least two distinct subgroups of index two accounting for those two subfields. That should help. Does it?
16h
comment Polynomial over finite field
That statement about polynomials of degree $n$ holds for all over integral domains. The usual proof via polynomial factoring works. The generalization fails after that. For example commutativity of the ring is absolutely essential. Anyway, good for you to question the validity of familiar arguments in an unfamiliar setting. Even better to remember the ingredients of the said argument so that you can later check their validity in a more general setting!
17h
comment How to find a solution to the elliptic curve
@pushpen.paul I used Mathematica. The commands were first In[1]:= EC = ContourPlot[y^2 == x^3 - 7, {x, 0, 40}, {y, -20, 200}], then In[2]:= tangent= Plot[6x-11,{x,0,36}], and then In[3]:= Show[{EC,tangent}]. I don't know if Wolfram Alpha can be coerced to do the same.
17h
comment How to find a solution to the elliptic curve
Paradox: Apparently you do know implicit differentiation? Using that (or simply solving for $y$ and differentiating) you should be able to show that the tangent at $(2,1)$ has slope $6$. So the equation of the tangent is $y-1=6(x-2)$ or, equivalently $y=6x-11$. So at the points of intersection we have $$(6x-11)^2=x^3-7.$$ You can find the solution of that cubic easily enough because we know that $x=2$ must be a double root. The third root gives you the point of intersection. Or you can use the point-doubling formula like Timbuc did.
17h
comment The dimension and basis of the set $F = \{(a+3b,a-b,2a-b,4b)| a,b \in \mathbb{R}\}$
You should try and explain which parts are giving you trouble. Otherwise it looks a lot like you want somebody to do your linear algebra assignment. André already gave you a useful hint. Why don't you work on that and report what you can do with the hint! It may happen that before you are done squeezing the last drop out of that hint the question will have been put on hold. That is the regulars way of telling you that you should improve the question. Frankly, it doesn't look good at the moment. Should that happen it's not the end. Just edit in your own work, and it will be "reopened".
17h
revised The dimension and basis of the set $F = \{(a+3b,a-b,2a-b,4b)| a,b \in \mathbb{R}\}$
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17h
comment How to find a solution to the elliptic curve
It is possible that the tangent makes order three contact with the ellipctic curve, when there truly would not be another point of intersection. But then the tangent will go "through the curve". Here the tangent is "locally above the curve". Because far away from origin we have $y\approx x^{3/2}$, the elliptic curve is guaranteed to catch up with the tangent line and cross over it.
17h
comment How to find a solution to the elliptic curve
The tangent does intersect the curve later on. May be your viewing area was too small? But you can also work it out algebraically.
17h
comment How to find a solution to the elliptic curve
I was planning on answering, and wanted to first look at the picture. You beat me to solving and answering (+1, of course), so I took the liberty of adding the picture of the elliptic curve together with the tangent line at the point of interest. If you don't want your terse answer tarnished by something like that, please let me know, and I'll remove it.
17h
revised How to find a solution to the elliptic curve
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1d
comment Whether polynomials $(t-1)(t-2),(t-2)(t-3),(t-3)(t-4),(t-4)(t-6)$ are linearly independent.
Your reasoning is correct (and shows a bit more mathematical maturity than the alternative of finding a dependency relation). A belated +1 as the question was now bumped :-)
1d
revised How to solve $\cos(5\alpha + \pi/2) = \cos(2\alpha + \pi/8)$ for $a$?
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1d
comment How to solve $\cos(5\alpha + \pi/2) = \cos(2\alpha + \pi/8)$ for $a$?
If you look at either the graph of cosine or its definition using the unit circle, then you can "see/conclude" that $\cos x=\cos y$, if and only if either $x=y+n\cdot2\pi$ for some integer $n$, or $x=-y+n\cdot2\pi$ for some integer $n$. Your equation is of the form "cosine of something = cosine of something else", so this can be used. You should end up with two families of solutions. It may or may not be possible to easily combine the two families, I haven't checked. Lycka till!
2d
comment Factoring a polynomial of fourth degree with false roots: $x^4+4$
$$x^4+4=(x^4+4x^2+4)-4x^2=(x^2+2)^2-(2x)^2=\cdots$$
2d
comment About Plücker embedding
And welcome to the site!