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

Question

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?

Answer 1

Hint $\ $ You seek an isomorphism of $\:\mathbb F_{11}[\sqrt{-1}]\:$ and $\rm\:\mathbb F_{11}[5-\sqrt{-1}\:\!],\:$ since $\rm\:x^2+x+4\:$ has roots $$\rm\dfrac{-1\pm\sqrt{-15}}2 \:\equiv\: \dfrac{10\pm\sqrt{-4}}2\:\equiv\: 5\pm\sqrt{-1}\pmod{11}$$

The discriminant (mod squares) characterizes isomorphism classes of quadratic extensions.

Answer 2

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 $\Bbb F_p=\Bbb Z/\Bbb Zp$.

Answer 3

Hint: Prove that any element in the first field can be uniquely written as $ax+b+(x^2+1)$. Prove the same for other field with $(x^2+1)$ replaced by $(x^2+x+4)$, where $(\dots)$ denotes the ideal generated by that element.

Now prove that they have 121 elements each. See Dummit & Foote if you need more help. Then, by the theory of finite fields, they are isomorphic. If you don't know the theory of finite fields, then define a map from the first to the second by sending $ax+b+(x^2+1) \mapsto ax+b+(x^2+x+4)$. Check that this map is homomorphism, and since L.H.S is a field this map is injective. Since both have 121 elements this map is surjective.

Answer 4

Here's an explicit isomorphism $f:\mathbb{Z}_{11}[x]/\langle x^2 + 1\rangle\rightarrow \mathbb{Z}_{11}[x]/\langle x^2+x+4\rangle$ between the 2 fields. Abusing notation, I'm going to refer to an element in the domain by it's natural preimage in $\mathbb{Z}_{11}[x]$, and likewise in the range.

First note that $1$ is uniquely determined in a field, so we must send $1$ to $1$. Using additivitiy, $f(n) = n$ for any $n\in\mathbb{Z}/11\mathbb{Z}$. So, the only remaining question is what $f(x)$ will be. Notice that $x^2 = -1$, so $f(x)$ must be a squareroot of $-1$ in the other field.

Writing $f(x) = ax+b$, we get $$-1 = f(x)^2 = a^2 x^2 +2abx + b^2 = a^2(-x-4)+2abx + b^2 = (2ab-a^2)x + b^2-4.$$

Now, $a\neq 0$ (else $f$ is not injective), so we learn that $2b=a$ and $b^2-4 = -1$.

The second equation, $b^2-4 = -1$ has 2 solutions (mod 11), $b= 5$ and $b = 6 (=-5)$. We'll pick $b=5$ (the choice doesn't matter). Then since $2b=a$, we get $a = 10 = -1$.

Thus, we have $f(x) = -x + 5$.

Putting this altogether, we have $f(ax+b) = -ax + 5a + b$ defining our isomorphism.

To see $f$ is injective, assume $f(ax+b) = 0$. Then, since $-ax + 5a+b = 0$, we must have $-a = 0$, so $a=0$. Once we know $a=0$, $b=0$ follows. So $f$ is injective. To see it's surjective, notice $f(-ax + 5a+b) = ax + b$, so $f$ is surjective.

Finally, we check it's a homomorphism. We have \begin{align*}f((ax+b) + (cx+d)) &= -(a+c)x + 5(a+c) + (b+d)\\\ &= -ax + 5a +b + -cx + 5c + d\\\ &= f(ax+b) + f(cx+d).\end{align*}

We also have \begin{align*} f((ax+b)(cx+d)) &= f(acx^2 + (ad+bc)x + bd)\\\ &= f((ad+bc)x + bd-ac)\\\ &=-(ad+cb)x + 5(ad+bc) + bd-ac\end{align*}

while \begin{align*} f(ax+b)f(cx+d) &= (-ax + 5a +b)(-cx+5c+d)\\\ &=acx^2 -5acx -adx -5acx +25ac+5ad-bcx+5bc+bd \\\ &= ac(-x-4) +x(-10ac-ad-bc) + 25ac+5ad+5bc+bd \\\ &= x(-11ac-ad-bc) + (21 ac+5ad+5bc+bd) \\\ &= x(-ad-bc) + 5ad+5bc +bd-ac\end{align*}

so $f((ax+b)(cx+d)) = f(ax+b)f(cx+d)$. Thus, $f$ is the desired isomorphism.