[Math] Prove that the fields $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4 \rangle$ are isomorphic

abstract-algebrafield-theoryfinite-fields

I have been stuck in this problem for some time now.

Prove that $x^2+2$ and $x^2+x+4$ are irreducible over $\mathbb{Z}_{11}$. Also, prove further $\mathbb Z_{11}[x]/\langle x^2+1\rangle$
and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic, each
having $121$ elements.

The first part is easy to prove since there is no element of $\mathbb Z_{11}$ that satisfies either of the polynomials given in the question. However, proving that $\mathbb Z_{11}[x]/\langle x^2+1\rangle$ and $\mathbb Z_{11}[x]/\langle x^2+x+4\rangle$ are isomorphic has been a challenge for me. How do I proceed? Moreover, how do I show that the the fields $\mathbb {Z}_{11}[x]/\langle x^2+1\rangle$ and $\mathbb {Z}_{11}[x]/\langle x^2+x+4\rangle$ each have $121$ elements?

Best Answer

A polynomial $f(x)$ with coefficients in any field $K$ and of degree $\leq3$ is irreducible over $K$ if and only if it has no roots in $K$. The given polynomials have no roots in $K=\Bbb Z_{11}$ and are therefore irreducible over $\Bbb Z_{11}$.

A standard fact about the ring of polynomals $K[X]$ is that its ideals are always principal, i.e. generated by one element. A standard consequence is that the ideals generated by irreducible polynomials are maximal. In addition, a quotient of a ring by an ideal is a field if and only if that ideal is maximal.

This explains why the given quotients are fields.

If $f(X)$ is irreducible of degree $d$, a full set of representants of the quotient $K[X]/(f(X))$ is given by the set of polynomials of degree $\leq d-1$ (this is an easy exercise). It follows that if $K$ is a finite field with $q$ elements the said quotient has $q^d$ elements. This explains why the given quotients have $121=11^2$ elements.

Finally, the general theory of finite fields tells us that for any prime power $q=p^f$ there exists a field with $q$ elements, which is unique up to isomorphism. Basically, the uniqueness follows from the fact that a finite field with $q=p^f$ elements is made up of the roots of the polynomial $X^q-X$ in some chosen algebraic closure of the basic field $\Bbb F_p=\Bbb Z/\Bbb Zp$.

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