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Let \begin{align*} F = \frac{\mathbb{Z}[x,y]}{(5,x^2 - y, xy + x + 1)}. \end{align*} Prove that $F$ is a field. With what field is $F$ isomorphic?

I let the ideal $ I = (5, x^2 - y, xy + x + 1)$. I see that $x^2 - y + I = 0 + I$, so that $x^2 + I = y + I$. So working in this quotient ring, I can substitute the indeterminate $y$ for $x^2$. Also I have $5 + I = 0 + I$. So the coefficients will go only up to $5$. I can also see that $$ F \cong \frac{Z_5[x,y]}{(x^2 - y, xy + x + 1)}. $$ But then I still don't see with what this is isomorphic. I tried defining the map $$ \phi: \mathbb{Z}_5[x,y] \to \mathbb{Z}_5[x]: f(x,y) \mapsto f(x, x^2)$$ but the kernel of this does not equal $(x^2 - y, xy + x + 1)$.

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    You are on the right track. You're taking the quotient of $\mathbb F_5[X]$ by $X^3+X+1$. – 2017-01-17
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    Ok, that would give me $X^3 = -X - 1 = 4X + 4$ in the quotient ring. So I will have only polynomials up to quadratic degree. But still from this I don't know how to proceed. – 2017-01-17
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    Well, you need to review how to construct finite fields, then. The polynomial $X^3+X+1$ is irreducible over $\mathbb F_5$, so the quotient I described is a field, and it has $5^3$ elements. – 2017-01-17

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$\frac {\mathbb Z[x,y]}{(5,x^2-y,xy+x+1)}\cong \frac {\mathbb Z[x,x^2]}{(5,x^3+x+1)}= \frac {\mathbb Z[x]}{(5,x^3+x+1)} $ we have $x^3+x+1$ irreducible polynomial in $\mathbb Z_5$ then $(5,x^3+x+1) $ is maximal ideal in $\mathbb Z[x]$ then $F$ is field.

or $\frac {\mathbb Z[x,y]}{(5,x^2-y,xy+x+1)}\cong \frac {\mathbb Z_5[x,y]}{(x^2-y,xy+x+1)}\cong\frac {\mathbb Z_5[x]}{(x^3+x+1)} $ , We have $\mathbb Z_5$ is a field then $\mathbb Z_5[x] $ is PID and $x^3+x+1$ irreducible polynomial in $\mathbb Z_5[x]$ then $(x^3+x+1) $ is maximal ideal in $\mathbb Z_5[x]$ then $F$ is field.

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    I Have a final question. Suppose I'm asked to determine if the polynomial $x^3 + 3x + 2$ has a root in $GF(5^3)$ (the finite field of $125$ elements). I don't have to explicitly show what is the root, just to determine whether there is one. Is there some trick to figure this out? – 2017-01-18