Irreducibility of a two variables polynomial

Question

Let $P_n(x,y)=\sum_{0\leq i+j\leq n}x^iy^j$ be a family of two variables polynomials. I want to know if it is irreducible over $\mathbb{C}[x,y]$, for any $n \in \mathbb{N}$. I used Sagemath and for the small values of $n$ it is irreducible. But I don't know how to prove it.

Answer 1

We have $P_1(x,y)=1+x+y+xy=(x+1)(y+1)$, which is not irreducible. Furthermore $$ P_2(x,y)=(x^2 + x + 1)(y^2 + y + 1). $$

Answer 2

Here is how I would approach this in sage. Define a polynomial ring over a nice base field, define the first polynomial explicitly, and factor it:

sage: r.<x,y> = PolynomialRing(QQ,2)
sage: p1=1+x+y+x*y
sage: p1.factor()
(y + 1) * (x + 1)

So, definitely not irreducible. Maybe there is a pattern to the factorizations? I need a formula for pn:

sage: n=2; p = sum( x^i*y^j for i in range(n+1) for j in range(n+1)); p;p.factor()
x^2*y^2 + x^2*y + x*y^2 + x^2 + x*y + y^2 + x + y + 1
(y^2 + y + 1) * (x^2 + x + 1)

Ok, the formula for p works for p2, and it factors again. I think I already see the pattern, but why not ask sage to keep going:

sage: for n in range(10):
....:      p = sum( x^i*y^j for i in range(n+1) for j in range(n+1) )
....:      print(p.factor())
....:
1
(y + 1) * (x + 1)
(y^2 + y + 1) * (x^2 + x + 1)
(y + 1) * (x + 1) * (y^2 + 1) * (x^2 + 1)
(y^4 + y^3 + y^2 + y + 1) * (x^4 + x^3 + x^2 + x + 1)
(y + 1) * (x + 1) * (y^2 - y + 1) * (y^2 + y + 1) * (x^2 - x + 1) * (x^2 + x + 1)
(y^6 + y^5 + y^4 + y^3 + y^2 + y + 1) * (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
(y + 1) * (x + 1) * (y^2 + 1) * (x^2 + 1) * (y^4 + 1) * (x^4 + 1)
(y^2 + y + 1) * (x^2 + x + 1) * (y^6 + y^3 + 1) * (x^6 + x^3 + 1)
(y + 1) * (x + 1) * (y^4 - y^3 + y^2 - y + 1) * (y^4 + y^3 + y^2 + y + 1) * (x^4 - x^3 + x^2 - x + 1) * (x^4 + x^3 + x^2 + x + 1)

Hrm, more data almost made it more confusing. The lines for 1, 2, 4, 6 all look like the same pattern. I'll just ask sage if that pattern works:

sage: for n in range(10):
....:      p = sum( x^i*y^j for i in range(n+1) for j in range(n+1) )
....:      print(p == sum( x^i for i in range(n+1))*sum( y^j for j in range(n+1)))
....:
True
True
True
True
True
True
True
True
True
True

Yup, it looks like $$\sum_{i,j} x^i y^j = \left(\sum_i x^i\right)\left(\sum_j y^j\right)$$

That formula is used pretty heavily in combinatorics, especially the slightly rewritten:

$$\sum_{i,j} a_i b_j = \left( \sum_i a_i \right) \left( \sum_j b_j \right)$$

which is a way of writing down the distributive property. "How do you multiply two lists of numbers that have been added up? Everything in the first times, everything in the second."

The factors in the "more data = less clear" section are also important and are called the cyclotomic polynomials. Had we used the ring CC ${}=\mathbb{C}$ instead, then (Sage would given an error message, but once we worked around it...) it would have factored everything into $ (x-\zeta^i)(y-\zeta^j) $ where $\zeta$ is an $n$th root of unity. I'm not sure that would have been clearer.