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Consider a polynomial of one variable over $\Bbb C$: $$p(x)=a_0+a_1x+\cdots+a_nx^n,\quad a_i\in\Bbb C.$$ We know from the Fundamental Theorem of Algebra that there exists $c,\alpha_i\in\Bbb C$ such that $$p(x)=c(x-\alpha_1)\cdots(x-\alpha_n),$$ i.e. we can factor $p$ in linear terms.

Now, what about polynomials $p(x,y)$ in two variables?

Is it still true that we can factor $p(x,y)$ in linear terms? I.e. can we always write $$p(x,y)=c(\alpha_1x+\beta_1y+\gamma_1)\cdots(\alpha_nx+\beta_ny+\gamma_n)$$ for some $c,\alpha_i,\beta_i,\gamma_i\in\Bbb C$?

user26857
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123
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3 Answers3

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Let me add to Martin's perfect answer that a homogeneous polynomial $f(x,y)$ (=sum of monomials of the same degree) in two variables does factor into linear homogeneous factors in an essentially unique way, that is up to permutations of the factors and multiplication of the factors by constants. More explicitly: $$f(x,y) =\sum_{i+j=d} a_{ij}x^iy^j=\prod _{k=1}^d(u_ix+v_iy)$$ However this is no longer true if the number of variables is $\geq 3$.
For example the polynomial $x^2+y^2+z^2$ is irreducible i.e. has no non-trivial factorization.

Finally note that if a homogeneous polynomial has a factorization, then the factors must be homogeneous too. For example : $$x^3+y^3+z^3-3xyz=(x+y+z)(x+wy+w^2z)(x+w^2y+wz) $$ (where $w=e^{2i\pi/3} $)

Edit
By request of a great contributor to this site, here is an explanation of why a homogeneous polynomial $f(x,y)$ in two variables over an algebraically closed field can be factored into linear polynomials:
Just write $$f(x,y)=\sum _{i=0}^da_{i}x^iy^{d-i}=y^d\sum _{i=0}^da_i(\frac xy)^i=y^d\prod_{i=1}^d (u_i(\frac xy)+v_i)=\prod_{i=1}^d (u_ix+v_iy)$$ The penultimate equality results from the factorization of a univariate polynomial into degree one factors.
[The calculation above is true only if $a_d\neq0$, i.e. if $f$ is not divisible by $y$.
If $f$ is divisible by $y$, one must very slightly modify the above but $f$ is still a product of linear factors, some of them now being equal to $y$. For example $x^2y^2+y^4=y^2(x+iy)(x-iy)$ ]

Jose Brox
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Georges Elencwajg
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  • Dear @Georges, maybe you want to explain why the homogeneous polynomials of degree $\ge2$ in $\mathbb C[x,y]$ are reducible (if I've understood correctly what you said). – user26857 Oct 04 '15 at 14:59
  • Dear @user26857: yes you have (of course!) perfectly understood what I meant. I have added an edit with the requested explanation. – Georges Elencwajg Oct 04 '15 at 18:22
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    Thanks. (+1) Meanwhile I've found a topic about this: http://math.stackexchange.com/questions/891109/factoring-homogeneous-polynomials-in-two-variables – user26857 Oct 04 '15 at 18:24
  • Thank you for this very nice answer (+1). For future readers: The calculation in the edit can be summarized as dehomogenize the homogeneous polynomial $f$ and then use that $f$ is defined over an algebraically closed field. Finally, the slight modification needed in case that $y$ divides $f$ mentioned in the edit amounts to writing $f=y^k \cdot g$ where $k\in{1,\ldots,d}$ and $g$ is a homogeneous polynomial of degree $d-k$ which is not divisible by $y$. One can then perform the above calculation for $g$. – Max Demirdilek Mar 04 '23 at 13:55
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My candidate to fail is $xy+1$. If this is the product of linear terms, we can have at most two of degree one. So $$ xy+1=(ax+by+c)(dx+ey+f). $$ We need $ad=be=0$, to avoid the squares. If $a=0$, then $b\ne0$ and so $e=0$. We have $$ xy+1=bdxy+bfy+cdx+cf. $$ Then $ cf=1$, $bf=cd=0$. So $b=d=0$, making the equality impossible. The other case is similar.

Martin Argerami
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  • Right. So would the proper generalization be that $p(x,y)$ can be factored in terms of degree 1? i.e. $a+bx+cy+dxy$? – 123 Oct 04 '15 at 08:25
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    Can we say something at all about the factorization of $p(x,y)$? – 123 Oct 04 '15 at 08:35
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    @123 we can say the factorization into irreducibles is unique, but even over an algebraically closed field there are irreducible polynomials of any degree with more than one variable. – Matt Samuel Oct 04 '15 at 10:53
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To show that this is not the case, look at $x^2 + y^2 - 1 = 0$, which is true all along the unit circle, if your idea was true that we could factor down to "individual zero points", we would need to have infinite number of factors since there are infinitude of points on the unit circle. Also linear terms would not be possible, since the unit circle is curved and not a set of lines.

mathreadler
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