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we know that $\mathbb{C}$ is an algebraically closed field so every polynomial $f$ in $\mathbb{C}[x]$ can be represented as

$f=(x-a_{1})^{\alpha_{1}} ....(x-a_{n})^{\alpha_{n}} $

and the only irreducible polynomials are the constant polynomials and of polynomials of the form $g=x-a $ ,$a \in \mathbb{C} $

Let $f \in \mathbb{C}[x,y]$ be a polynomial

When do we say that $f$ is irreducible in $\mathbb{C}[x,y]$? Do the irreducible polynomials have an exact form like in the case of $\mathbb{C}[x]$? if not how we can check whether $f$ is irreducible or not ?

For example is this polynomial irreducible in $\mathbb{C}[x,y]$ or not

$h(x, y)=x^{p-1}+x^{p-2}+....+x+1-y$

Aster Phoenix
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    We often can use Eisenstein. See also this post and other ones here. – Dietrich Burde Jan 05 '21 at 10:21
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    One way to do things, is that $\mathbb C[x,y] \cong (\mathbb C[x])[y]$ , so a factorization of this polynomial in $\mathbb C[x,y]$ is akin to a factorization of it in $(\mathbb C[x])[y]$. Looking at it from that viewpoint, your polynomial is of "degree one" in $y$, which we know to be irreducible. Using this correspondence, you can also formally attempt to prove that the given polynomial is irreducible, by imagining what a factorization would look like in terms of powers of $y$. Please look up the numerous posts on "irreducibility of polynomials" on MSE for better intuition on the topic. – Sarvesh Ravichandran Iyer Jan 05 '21 at 10:24
  • @teresa Lisbon can you give me an example to make the idea more clear i mean view polynomial in $(\mathbb{C}[x])[y]$ – Aster Phoenix Jan 05 '21 at 10:30
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    For example, $$xy + 3y^2 + 2x + 1 + 4y^3 + 3x^2y + 4y^2x^7\ = (2x+1) + y(x + x^2) + y^2(3 + 4x^7) + y^3(4)$$ , where the left side is the polynomial in $x,y$ and the right side is a polynomial in $y$ with coefficients which are polynomials in $x$. – Sarvesh Ravichandran Iyer Jan 05 '21 at 10:35
  • @Dietrich Brude can you give an example of using Eisenstien method in $\mathbb{C[x,y]}$ – Aster Phoenix Jan 05 '21 at 10:35
  • @Teresa Lisbon but in your example is your polynomial irreducible or not ? – Aster Phoenix Jan 05 '21 at 10:38
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    Not sure, my polynomial is randomly constructed just to give you an idea. How would I check it? It is a cubic polynomial in $y$, so one of the factors , if it is there is of the form $y - p(x)$ for a polynomial $p$ in $x$. So put $y = p(x)$ and then see if it works? It feels like the polynomial is irreducible, the reason being that too many powers of $x$ are missing between $2$ and $7$, but this is only heuristic, and not indicative of rigorous thinking. – Sarvesh Ravichandran Iyer Jan 05 '21 at 10:55

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The definition of irreducible polynomials $\mathbb{C}[x,y]$ is the same as in general commutative rings: A polynomial $f$ is irreducible if

  • it is not a unit or $0$, i.e. constant in the case $\mathbb{C}[x,y]$
  • if there is a factorization $f=ab$ then $a$ or $b$ has to be a unit.

I don't know if there is some kind of exact form for irreducible polynomials over $\mathbb{C}[x,y]$, the generalization of the fact that the irreducibles in $\mathbb{C}[x]$ are exactly those of the form $x-a$ is the Nullstellensatz concerning maximal ideals, but not single polynomials.

For your example: $h(x,y)=x^{p-1}+\dots+x+1-y$. This is irreducible since it has degree 1 in $y$.

leoli1
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