Q. ( y^{2} + x^{4} y + x + 1, f_{2}[x,y] ).

Q: Is this polynomial y^2 + x^4 y + x + 1 considered in \mathbb{F}_2[x,y]?

Yes. The notation f_2[x,y] means the polynomial ring over the field with two elements, so coefficients are mod 2 and arithmetic is in characteristic 2.

Q: Can it be factored in \mathbb{F}_2[x,y]?

Treat it as a quadratic in y. It factors iff there exist u,v\in\mathbb{F}_2[x] with u+v=x^4 and uv=x+1. Test by searching linear factors y+g(x) or use computer algebra to decide irreducibility.

Q: How do I find roots for y over the rational function field \mathbb{F}_2(x)?

The equation is quadratic in y; its roots lie in a degree-2 extension of \mathbb{F}_2(x). Solve y^2+x^4y+(x+1)=0 in that extension or use a CAS to compute the explicit algebraic expressions for the two roots.

Q: What are the affine \mathbb{F}_2-rational points (x,y\text{ in }\{0,1\})?

Evaluate all four pairs: for x=0 we get y^2+1 so y=1. For x=1 we get y^2+y so y=0,1. Points: (0,1),(1,0),(1,1).

Q: Is the plane curve singular?

Compute partials: f_x = 4x^3y+1 = 1 (mod 2) and f_y = 2y+x^4 = x^4. Since f_x is the constant 1, there are no affine singular points (no simultaneous vanishing). Check points at infinity separately.

Q: What is the total degree and expected genus of the projective closure?.

The highest total degree term is x^4y, so degree d=5. Smooth projective plane curve of degree 5 has genus (d-1)(d-2)/2=6. Verify smoothness at infinity to confirm this genus.

Q: How can I check factorization or properties using software?

Use a CAS with characteristic 2: e.g. Sage: factor(y^2 + x^4*y + x + 1, modulus=2); Magma/Maple/Mathematica have analogous factor or FactorMod commands with modulus 2.

Answer

Let \(f=y^2+x^4y+x+1\in\mathbb{F}_2[x,y]\).

If \(f\) were reducible, then because it is quadratic in \(y\), it would factor as:

\(f=(y+a)(y+b)\)

where \(a,b\in\mathbb{F}_2[x]\).

Expanding gives:

\((y+a)(y+b)=y^2+(a+b)y+ab\)

Compare this with:

\(y^2+x^4y+x+1\)

So, we must have:

\(a+b=x^4\)

and:

\(ab=x+1\)

Since \(b=x^4+a\), substitute this into \(ab=x+1\).

\(a(x^4+a)=x+1\)

Expand the left side.

\(ax^4+a^2=x+1\)

Over \(\mathbb{F}_2\), addition and subtraction are the same, so this can be written as:

\(a^2+x^4a+x+1=0\)

Now write \(a\) as a polynomial:

\(a=\sum c_i x^i\)

Then \(a^2\) contains only even-degree monomials because squaring doubles every exponent.

The term \(x^4a\) contains monomials whose degrees are each at least \(4\) higher than the degrees in \(a\).

Therefore, \(a^2+x^4a\) cannot produce a monomial of degree \(1\).

But the expression \(x+1\) contains the degree-\(1\) term \(x\).

This is impossible, so no polynomial \(a\in\mathbb{F}_2[x]\) can satisfy the equation.

Therefore, \(f\) does not factor over \(\mathbb{F}_2[x,y]\).

Final result: \(f\) is irreducible in \(\mathbb{F}_2[x,y]\).

Detailed Explanation

We work in the polynomial ring F2[x,y] (coefficients modulo 2). Let

\[f(x,y)=y^{2}+x^{4}y+x+1.\]

Step 1 — form of a nontrivial factorization in y.

If f is reducible in F2[x,y], since deg_y(f)=2 it must factor as a product of two linear factors in y. Write the most general possibility

\[\big(a(x)\,y+b(x)\big)\big(c(x)\,y+d(x)\big),\]

with a(x),b(x),c(x),d(x)\in F_{2}[x]. Expanding gives

\[a(x)c(x)\,y^{2}+\big(a(x)d(x)+b(x)c(x)\big)y+b(x)d(x).\]

But the coefficient of y^2 in f is 1, so a(x)c(x)=1 in F2[x]. The only units in F2[x] are 1, hence a(x)=c(x)=1. Therefore any factorization (if it exists) must have the shape

\[\big(y+A(x)\big)\big(y+B(x)\big)\]

with A(x),B(x)\in F_{2}[x]. Expanding yields the system

\[A(x)+B(x)=x^{4},\]

\[A(x)B(x)=x+1.\]

Step 2 — degree argument and contradiction.

Take degrees (over x). From A·B = x+1 we have deg(A)+deg(B)=deg(x+1)=1. Thus one of A,B has degree 1 and the other has degree 0 (a constant). The only constants in F2 are 0 and 1; neither factor can be 0 because their product is x+1≠0, so the constant must be 1. Hence, up to order, the only possible pair is

\[A(x)=1,\quad B(x)=x+1\]or

\[A(x)=x+1,\quad B(x)=1.\]

Compute A+B in either case:

\[1+(x+1)=x,\]

which is not equal to x^4. This contradicts A+B=x^4.

Step 3 — conclusion.

Because no polynomials A(x),B(x)\in F_{2}[x] satisfy the required relations, f cannot factor as a product of two linear-in-y polynomials in F2[x,y]. Therefore f is irreducible in F2[x,y].

Final answer: The polynomial \[y^{2}+x^{4}y+x+1\] is irreducible in F_{2}[x,y].

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Algebra FAQs

Is this polynomial \(y^2 + x^4 y + x + 1\) considered in \(\mathbb{F}_2[x,y]\)?

Yes. The notation \(f_2[x,y]\) means the polynomial ring over the field with two elements, so coefficients are mod \(2\) and arithmetic is in characteristic \(2\).

Can it be factored in \(\mathbb{F}_2[x,y]\)?

Treat it as a quadratic in \(y\). It factors iff there exist \(u,v\in\mathbb{F}_2[x]\) with \(u+v=x^4\) and \(uv=x+1\). Test by searching linear factors \(y+g(x)\) or use computer algebra to decide irreducibility.

How do I find roots for \(y\) over the rational function field \(\mathbb{F}_2(x)\)?

The equation is quadratic in \(y\); its roots lie in a degree-2 extension of \(\mathbb{F}_2(x)\). Solve \(y^2+x^4y+(x+1)=0\) in that extension or use a CAS to compute the explicit algebraic expressions for the two roots.

What are the affine \(\mathbb{F}_2\)-rational points \((x,y\text{ in }\{0,1\})\)?

Evaluate all four pairs: for \(x=0\) we get \(y^2+1\) so \(y=1\). For \(x=1\) we get \(y^2+y\) so \(y=0,1\). Points: \((0,1),(1,0),(1,1)\).

Is the plane curve singular?

Compute partials: \(f_x = 4x^3y+1 = 1\) (mod 2) and \(f_y = 2y+x^4 = x^4\). Since \(f_x\) is the constant 1, there are no affine singular points (no simultaneous vanishing). Check points at infinity separately.

Is the polynomial square-free (no repeated factor)?

What is the total degree and expected genus of the projective closure?\(.\)

The highest total degree term is \(x^4y\), so degree \(d=5\). Smooth projective plane curve of degree \(5\) has genus \((d-1)(d-2)/2=6\). Verify smoothness at infinity to confirm this genus.

How can I check factorization or properties using software?

Use a CAS with characteristic 2: e.g. Sage: factor(\(y^2 + x^4*y + x + 1\), modulus=2); Magma/Maple/Mathematica have analogous factor or FactorMod commands with modulus 2.

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