# Thread: Algebraic geometry---> irreducible polynomial

1. ## Algebraic geometry---> irreducible polynomial

Hello, I am trying to show that y^5 -x^2 is irreducible in R[x,y]. (R represents the set of real numbers and R[x,y] , the polynomial ring in two variables)

I have tried everything but still don't know how to show this!

Here's what I tried:

y^5 -x^2 = fg , where f&g are non constant polynomials in R[x,y].

Then let x=t^5 and y=t^2.

so, 0=f(t^5,t^2)g(t^5,t^2), which implies

either f(t^5,t^2)=0 or......

My professor said this is the best approach and I should find a contradiction.

Any help will be grateful thanks.

2. Originally Posted by Algebraicgeometry421
Hello, I am trying to show that y^5 -x^2 is irreducible in R[x,y]. (R represents the set of real numbers and R[x,y] , the polynomial ring in two variables)

I have tried everything but still don't know how to show this!

Here's what I tried:

y^5 -x^2 = fg , where f&g are non constant polynomials in R[x,y].

Then let x=t^5 and y=t^2.

so, 0=f(t^5,t^2)g(t^5,t^2), which implies

either f(t^5,t^2)=0 or......

My professor said this is the best approach and I should find a contradiction.

Any help will be grateful thanks.
look at $-x^2 + y^5$ as an element of $D[x],$ where $D=\mathbb{R}[y].$ note that the units of $D$ are exatly the units of $\mathbb{R}.$ suppose that $-x^2+y^5=f(x)g(x),$ for some non-units $f,g \in D.$ then

$\deg f(x)=\deg g(x)=1,$ i.e. $f(x)=ax+b, \ g(x)=cx+d,$ where $a,b,c,d \in D.$ now $-x^2+y^5=(ax+b)(cx+d)$ will give us $ac=-1, \ ad+bc=0, \ bd=y^5$ and so $(ad)^2=y^5,$

which is obviously impossible because the degree of $(ad)^2 \in D$ is an even number.

3. ## Thanks

Thank you that helps a lot.