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$ y^3 − (x^2)y + x(y^2) − 3x$ in $F5[x, y].$

Is it right to go through all the $(x,y)$ from $0,1,2$ and $4$

but when $x=0,y=0$ and $x=0,y=4$ and some other situations that the polynomial can be $0,$ does it mean that polynomial is reducible

gymbvghjkgkjkhgfkl
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vivi.wang
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  • An irreducible polynomial can have zeros. Simple example: x + y. Irreducible means the polynomial can't be expressed as a product of two nonconstant polynomials. – quasi Mar 09 '17 at 09:23

2 Answers2

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Suppose $f = y^3-x^2y+xy^2-3x$ is reducible in $F_5[x,y]$.

Thus, assume $f = gh$, for some non-constant $g,h \in F_5[x,y]$.

We can regard

$$f = (-y)x^2 + (y^2 - 3)x + (y^3)$$

as a quadratic in $x$, with coefficients in $F_5[y]$.

Then, since the coefficients $\,(-y)$, $\,(y^2-3)$, $\,(y^3)\,$ have no common factor in $F_5[y]$, it follows that $g,h \notin F_5[y]$, hence $g(x,y),h(x,y)$ must each have degree $1$ in the variable $x$.

Then \begin{align*} &f(x,y) = g(x,y)h(x,y) \text{ in }F_5[x,y]\\[6pt] \implies\; &f(x,-1) = g(x,-1)h(x,-1) \text{ in }F_5[x]\\[6pt] \implies\; &x^2 - 2x - 1 = g(x,-1)h(x,-1) \text{ in }F_5[x]\\[6pt] \end{align*}

Since $g(x,y),h(x,y)$, as elements of $F_5[x,y]$, have degree $1$ in $x$, the polynomials $g(x,-1),h(x,-1)$, as elements of $F_5[x]$, have degree at most $1$. But since the product in $F_5[x]$ of $g(x,-1)$ and $h(x,-1)$ is $x^2 - 2x - 1$, it follows that each of $g(x,-1),h(x,-1)$ must have degree exactly $1$, hence

\begin{align*} &x^2 - 2x - 1 = g(x,-1)h(x,-1) \text{ in }F_5[x]\\[6pt] \implies\; &x^2 - 2x - 1 \text{ is reducible in }F_5[x]\\[6pt] \implies\; &\text{The discriminant }D\text{ of }x^2 - 2x - 1\text{ is a square in }F_5\\[6pt] \end{align*}

contradiction, since

$$D = (-2)^2 - 4(1)(-1) = 8 = 3\qquad\qquad\qquad\;\;\,$$

which is not a square in $F_5$.

It follows that $f$ is irreducible in $F_5[x,y]$.

quasi
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Consider $f$ as polynomial in $y$ over the unique factorization ring $F_5[x]$.

A straightforward generalization of Eisenstein's Criterium holds for such rings.

Now, using the prime element $x\in F_5[x]$ and applying Eisenstein's Criterium, we get that there is no factorization $f=g h$, with $g$ and $h$ polynomials of $F_5[x][y]$ having both degrees strictly less than the degree of $f$. Since $f$ is a monic polynomial, it is irreducible in $F_5[x][y]$.

Moreover, $f$ is also irreducible in the polynomial ring $F_5(x)[y]$,where $F_5(x)$ is the ring of rational functions of $F_5[x]$.

texavatar
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