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Picture above ^^

Hello I am really needing for someone to help for me to understand the following;

I know that in fields, quadratics and cubics are reducible if and only if they have roots. My question is, what is the form of these roots.

For example to explain my confusion, consider the field with 121 elements, ie , with coefficients in $\mathbb{Z}/11\mathbb{Z}$ and quotient with an ideal of a polynomial of degree two.

$\mathbb{L}=F_{11}[X] \setminus (p(x))$ where $p(x)$ is a irreducible polynomial of degree two.

When we ask if we can solve the polynomial $t^{2}-2$ in this field, are we asking about t coming from $\mathbb{Z}/11\mathbb{Z}$ and $t^{2}$ from $\mathbb{Z}/11\mathbb{Z}$ respectively, or


Are we asking if there is a polynomial $(ax+b)$ where $a, b$ $\mathbb{Z}/11\mathbb{Z}$ such tht $(ax+b)^{2}-2=0$ ie our t would be one of our 121 elements in the field and so would $t^{2}$

I am leaning towards this one.

I think I am confused with all the x, because they are really not the same , for example If I was asked is $t^2-2$ solvable, would that be equivalent but less confusing because I am considering t to be any element of the finite field?

Hope this makes sense, I really need to be able to understand this.

Thank you everyone


I am still confused and starting to feel like I don't know if I ever will be able to understand this or even articulate my question, I am so confused with how we use x to make a polynomial of degree 2 in order to get a field of 121 elements , and then are asked to solve polynomials that are in terms of x as well.

in the picture for example it says construct the field with 121 elements.

How? How can I know which polynomial of degree two are reducible or not. Reducible or irreducible WHERE? in Z/11?

Is the method of listing the squares mod 11 the way to determine if $x^2-c$ has a root and is irreducible??

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    $\begingroup$ OK - first of all $\mathbb Z/(10)$ is not a field, because $2\cdot 5 = 0$... Did you mean $\mathbb Z/(11)$? $\endgroup$ – peter a g Dec 10 '15 at 2:31
  • $\begingroup$ Yes that was a typo, thanks I updated it $\endgroup$ – Quality Dec 10 '15 at 2:41
  • $\begingroup$ "in fields, quadratics and cubics are irreducible if and only if they have roots." I think you mean they do not have roots in the field. $\endgroup$ – Umang Dec 10 '15 at 2:58
  • $\begingroup$ You really need to clarify your definition of the field with $121$ elements. What you've written makes almost no sense, but I gather that you want it to be $(\mathbb{Z}/11) [X]/(p(X))$ for some quadratic $p$. What is $p$? $\endgroup$ – Slade Dec 10 '15 at 3:00
  • $\begingroup$ Yes apparently the polynomial t^{2}-2 will have roots in this field regardless of which p we choose, as long as it is irreducible of degree 2 $\endgroup$ – Quality Dec 10 '15 at 3:03

First, an example, where $p(X)=X^2-6$.

$X^2-6$ is irreducible of degree $2$ in $\mathbb{Z}/11[X]$, so the quotient $k = (\mathbb{Z}/11)[X]/(X^2-6)$ is a field with $121$ elements.

We can now ask whether $T^2-2$ is irreducible in $k[T]$. To check if it has a root, we can take an arbitrary coset $\overline{aX+b}\in k$, with $a,b\in\mathbb{Z}/11$, and plug it in to $T^2-2$.

We get $(\overline{aX+b})^2 - 2 = a^2\overline{X}^2 + 2ab \overline{X} + b^2 - 2$. Since $\overline{X}^2 = 6$, this is $2ab\overline{X} + (6a^2 + b^2 - 2)$.

Since this has to equal $0$, and the vectors $1$ and $\overline{X}$ are independent over $\mathbb{Z}/11$, we must have $2ab = 0$ and $6a^2 + b^2 - 2 = 0$. Either $a=0$, in which case $b^2 - 2 = 0$, which has no solutions in $\mathbb{Z}/11$, or $b=0$, giving us $6a^2 - 2 = 0$. This has solutions $a = \pm 2$, so the roots are $\pm 2 \overline{X}$.

We can do a similar exercise if $p(X)$ is any irreducible quadratic, but we will need to use its irreducibility somewhere in the proof.

We don't necessarily have to use brute force like this if we know some theory. Every field with $121$ elements is isomorphic, which means that every quadratic splitting field over $\mathbb{Z}/11$ is isomorphic, which means that every quadratic is reducible in $\mathbb{Z}/11$.

We could also approach it like this: it is a well-known fact that the multiplicative group of a finite field is cyclic. So the group $k^\times$ is cyclic of order $120$, and it includes, as the unique cyclic subgroup of order $10$, the group $(\mathbb{Z}/11)^\times$. This implies that each element of $(\mathbb{Z}/11)^\times$ is a square in $k^\times$.

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  • $\begingroup$ Thank you, so by this, would not $p(x)=x^{2}-2$ also by irreducible in Z11 and hence Z11/(x^2-2) would be a field? PS: I will add a picture of the question $\endgroup$ – Quality Dec 10 '15 at 5:12
  • $\begingroup$ Yes, but the fields $\Bbb F_{11}[X]/(X^2-6)$ and $\Bbb F_{11}[X]/(X^2-2)$ are isomorphic, even if not canonically. $\endgroup$ – Lubin Dec 10 '15 at 5:41
  • $\begingroup$ @Quality Oh, if you get to pick the presentation of $k$, then $p(X)=X^2+1$ is probably simplest for calculation purposes. $\endgroup$ – Slade Dec 10 '15 at 5:47
  • $\begingroup$ Im not sure I follow, So $x^{2}-2$ is irreducible in $Z11 $and its a Z11/(f(x)) , but x^{2}-2 is reducible in Z11/(x^{2}-2)? $\endgroup$ – Quality Dec 10 '15 at 5:53
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    $\begingroup$ @Quality You really have to stop using $x$ to mean multiple things. Yes, the polynomial $X^2-2$ is irreducible in $(\mathbb{Z}/11)[X]$. As a consequence, $k=(\mathbb{Z}/11)[X]/(X^2-2)$ is a field with $121$ elements. The polynomial $T^2-2$ is then reducible in $k[T]$; in fact, the coset $\overline{X}$ is a root, precisely because $\overline{X^2 - 2} = \overline{0}$. $\endgroup$ – Slade Dec 10 '15 at 6:14

It can get confusing, no doubt about it. But to hold your confusion to a minimum, you should decide once and for all how you are going to represent your elements of $\Bbb F_{121}$. Since $-1$ is not a square modulo $11$, I would have chosen to write $\Bbb F_{121}$ as $\Bbb F_{11}(i)=\Bbb F_{11}(\sqrt{-1}\,)=\Bbb F_{11}[X]/(X^2+1)$. This has the merit of looking familiar. Let’s do some computations:

You know that $X^2-2$ is irreducible over $\Bbb F_{11}$, but I’ll point out to you how to find its factorization over $\Bbb F_{11}(i)$. Notice that $-1$ and $2$ both are nonsquares, so their quotient, $-1/2$, must be a square modulo $11$, and you see that $2/(-1)=-2\equiv9=3^2\pmod{11}$, so that you can take $\sqrt2=3\sqrt{-1}=3i$. This means that although $X^2-2$ is irreducible as a $\Bbb F_{11}$-polynomial, it’s reducible as a $\Bbb F_{121}$-polynomial, namely $X^2-2=(X-3i)(X+3i)$. Similarly, $6/(-1)=-6\equiv5\equiv16=4^2\pmod{11}$, so that you may take $\sqrt6=4i$ and factor $X^2-6$ over $\Bbb F_{121}$ as $(X-4i)(X+4i)$.

Remember that once you’ve chosen how you’re going to represent your non-prime finite field $\Bbb F_{p^n}$ as $\Bbb F_p[X]/(P(X))$, you’ll use your $\Bbb F_p$-irreducible polynomial $P(X)$ of degree $n$ throughout your computations, never changing it. If, in midstream, you wish to exchange your horse $P$ for another horse $Q$, the process may be very difficult, and lead you into all sorts of problems.

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  • $\begingroup$ Thank you, PS: How can I know the quotient of two non squares must be a square? $\endgroup$ – Quality Dec 11 '15 at 1:15
  • $\begingroup$ Follows from the fact that the multiplicative group of a finite field is cyclic. $\endgroup$ – Lubin Dec 11 '15 at 2:42

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