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Suppose my ring is $\mathbb{R}[x,y]$, and the polynomial in question in $x^2+y^2$.

Would this be a proof that its irreducible: think of $x^2+y^2$ as a polynomial in $x$ with coefficient in $y$, ie $(1)x^2+(0)x^1+(y^2)x^0$ and use determinant to see if it has real roots, so $\Delta=0^2-(4*1*y^2) = -4y^2$, since $y$ is in the reals $\Delta<0$ so $x^2+y^2$ is irreducible in $\mathbb{R}[x,y]$.

If it is a proof, would it also be true that $x^2+y^2$ is reducible in $\mathbb{C}[x,y]$.

Finally, how would it work with $\mathbb{R}[x,y,z]$?

KReiser
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1 Answers1

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It is perhaps easier to say that $$x^2+y^2=(x+iy)(x-iy)$$ is the factorization into irreducible elements of $x^2+y^2$ in $\mathbb C[x,y]$. Obviously $x^2+y^2$ cannot be factored in $\Bbb R[x,y]$.

On the other hand, $x^2 +y^2 + z^2$ is already irreducible in $\mathbb C [x,y,z]$.

$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$

Dietrich Burde
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