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?

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    $\begingroup$ $(x+6)^2+1=x^2+x+4$ holds in $\mathbb Z_{11}[x]$ $\endgroup$ – Martin Sleziak Apr 21 '12 at 18:05
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    $\begingroup$ From a purely abstract point of view, both quotients are fields of $121$ elements (since both polynomials are irreducible), hence both are splitting fields of $x^{121}-x$ over $\mathbb{Z}_{11}$, hence they are isomorphic. $\endgroup$ – Arturo Magidin Apr 21 '12 at 22:10

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.

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  • $\begingroup$ +1 Note that a trick like this will ALWAYS work in proving that two fields of cardinality $p^2$ are isomorphic (see also Martin Sleziak's comment above). We get such fields by joining a root of a quadratic to the prime field. The discriminant of that quadratic has to be a non-square. But because the multiplicative group $\mathbb{Z}_p^*$ is cyclic, we see that the ratio of any two non-squares is a square, so if you join the square root of one non-square, you join them all! Of course, the result in Arturo's comment to the OP settles the question even in the more general case. $\endgroup$ – Jyrki Lahtonen Apr 22 '12 at 17:32

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.

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  • $\begingroup$ Let me ask you one question: If $a=0$ then why $f$ is NOT injective? $\endgroup$ – ZFR Jun 8 '18 at 21:08
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    $\begingroup$ If $a = 0$, then $f(x) = f(b) = b$, but $x\neq b$. $\endgroup$ – Jason DeVito Jun 9 '18 at 3:12
  • $\begingroup$ Your solution is nice and well-written in simple language! But some moments are unclear to me. I have realized that $f(1)=1$ and by additivity $f(n)=n$ for any $n\in \mathbb{Z}_{11}$. And you writing that "the only remaining question is what $f(x)$ will be?" Since you have derived that $f(n)=n$ why you cannot conclude that $f(x)=x$ (this moment confuses me a lot)? Instead of that you suppose that $f(x)=ax+b$ and then through equations you found values of $a$ and $b$. Can you clarify this moment, please? Would be very grateful for that! $\endgroup$ – ZFR Jun 9 '18 at 8:01
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    $\begingroup$ @RFZ: $x$ is a specific element in $\mathbb{Z}_{11}[x]/\langle x^2 + 1\rangle$ and it is not obtained in the sequence $0, 1, 2,3,..., 10$. Hence, the fact that $f(0) = 0$ and $f(1) = 1$ and ... and $f(10) = 10$ in no way forces $f(x) = x$. And, as my answer demonstrates, the choice $f(x) = x$ is incompatible with $f$ being a field homomorphism. $\endgroup$ – Jason DeVito Jun 10 '18 at 23:55

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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  • $\begingroup$ Wonder what I said so wrong to deserve a negative .... $\endgroup$ – Andrea Mori Apr 21 '12 at 18:57
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    $\begingroup$ I was the one who down voted you. Sorry about that. I did so because I didn't think that you answered enough of what was asked in the original question. You say that the polynomials are irreducible, but that doesn't seem to be the main question. Why are there 121 elements? Why is the quotient a field? (I will upvote you if you answer these questions...) $\endgroup$ – Thomas Apr 21 '12 at 20:06
  • $\begingroup$ @Thomas, I find a bit strange that somebody asking this kind of a question for a finite field does not have enough mathematical background to understand what I wrote without needing further details. Anyway, I will edit my answer adding some details. $\endgroup$ – Andrea Mori Apr 21 '12 at 21:30
  • $\begingroup$ Thanks. Much better answer. $\endgroup$ – Thomas Apr 21 '12 at 21:52

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.

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    $\begingroup$ The map you describe is not a homomorphism since $f(\overline{x}*\overline{x}) = f(\overline{x}^2) = f(-\overline{1}) = -\overline{1}$ while $f(\overline{x})*f(\overline{x}) = \overline{x}\overline{x} = -\overline{x}-\overline{4}$. (I'm using overlining to denote the images of elements in the quotient fields). $\endgroup$ – Jason DeVito Apr 21 '12 at 21:08

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