Question

Algebraic numbers to HS students.

Answer 1

“as elements of a polynomial ring over an ideal”

For real? High schools students? If you want to teach them some abstract algebra, why not just do a little group theory, motivate it with symmetry, etc.  I just cannot imagine talking about rings and ideals as someone’s first exposure to abstract math.

Answer 2

I'm not a teacher, but here's my intuition for algebraic numbers.

You start with the natural numbers. You introduce the integers as all solutions x to x+y=z for any and all natural numbers y and z, which also completely solves the equation  x+w=0 for all integers w. Rationals come from the similar equations x*y=z and px+q=0 for integers y,z and rationals p,q. The constructible numbers are the solutions to quadratic equations with integer coefficients. Each extension has been to solve some polynomial with integer coefficients, firstly linear polynomials, and then quadratic. Algebraic numbers are then the natural conclusion when extending to polynomials of arbitrary degree with integer coefficients.

Again, I am not a teacher. This might not be the best way to teach these students, but it's how I self-taught myself the intuition for algebraic numbers.

Answer 3

>I want to bring them through the construction of algebraic numbers as elements of quotients of a polynomial ring over an ideal.

This seems well outside the scope of your audience. Is there a specific reason you want to formally introduce algebraic numbers to them?

>I do not want to just say that algebraic numbers are the solutions of polynomial equations

This seems much more appropriate.

Answer 4

The thing is that a number is algebraic over a ring R if its the solution of a polynomial over that ring. So there is no way around this. Quotienting R[x] by ideals simply "adds" these elements to R.

Perhaps give examples and non examples of algebraic extensions and then show how to construct them using quotient rings. Good examples are Q[i], Q[sqrt(2)] and good non examples are Q[x] and Q[e]. I would say, stick to Q and Z for examples and not rings with positive characteristic.

Answer 5

imo quotients of rings by ideals really make more sense in the context of modules, and more specifically in the category of modules over a commutative ring. maybe start with abelian categories in general and then at least state the embedding theorem?

Answer 6

Perhaps you can spark some interest in a couple lessons,  by using some heavy hitting but intentionally unproven theorems to "prove" something fun and counterintuitive about algebraic numbers, like the idea that we can't write them all as arithmetic expressions with radicals (since not all quintics can be solved).

I don't know how feasible that would be (because I never took algebra or Galois theory), but if you can make it into a fun and surprising adventure, then I bet mathy students would enjoy the story, even if they don't get the deeper stuff!

Answer 7

I'd say introduce modular arithmetic as quotients of the form x + nZ, show that addition and multiplication still work, then try to show you can use it to make a space of numbers where a polynomial has a solution.

Answer 8

Start with the complex numbers. Remind them how `i` is constructed by freely adjoining a square root of `-1`, then generalize. So for example, do the dual numbers next. Then start doing it over the integers instead of the reals and show how things are different and how there are more possible roots to join. Then demonstrate how a quotient can be used to force an equation to hold. I wouldn't introduce ideals or even full ring theory at first.

Answer 9

Take this from someone who took Calculus II and III at a local university while in high school....you aren't teaching this concept to high schoolers.

Answer 10

Hi,

 

You already got a lot of good answers, but here is one. Define rational linear independence by saying that two real (or complex) numbers  x,y  are linearly independent if all the expressions like  4x+7y  or   4/3 x + 8/5 y  are different from each other. So for instance 1 is linearly independent of the square root of 2 because otherwie q.1 + r.\sqrt{2} = w.1 + u.\sqrt{2}  and you could solve to get the square root of 2 to equal a rational number unless q=w and r=u. And extend this definition to more than sequences of length 2.

 

Now you say a number is transcendental if all its powers are linearly independent (over the rationals). That it is very hard to prove this but for instance \pi and e are known to be transcendental. But any number that is not transcendental is called 'algebraic.'

 

The reason the square root of 2 is algebraic is that its second power is two times its zero'th power.