Prove that if $t$ is a root of $f(x)$, then $t^2$ is a root of $f(x)$, where $f(x)=x^n+x^{n-1}+\cdots+x+1 \in \Bbb{Z}_2[x]$

Suppose $$f(x)=x^n+x^{n-1}+x^{n-2}+\cdots +x+1$$ is a polynomial of degree $$n$$ defined over $$\Bbb{Z}_2[x]$$. When $$a$$, $$b \in \Bbb{Z}_2$$, we have $$(a + b)^2 = a^2 + b^2$$. This idea can be extended to any number of variables, that is, $$(a_1+a_2+\cdots+a_r)^2=a_1^2+a_2^2+\cdots+a_r^2$$, where $$a_i \in \Bbb{2}$$. Use this fact to help prove that if $$t$$ is a root of $$f(x)$$, then $$t^2$$ is a root of $$f(x)$$.

I understand that in $$\mathbb{Z}_2$$, the only elements are $$0$$ and $$1$$. $$0^2=0$$ and $$1^2=1$$. Is this all I need to prove that if $$t$$ is a root, so is $$t^2$$? Or am I missing something?

• What’s the question? – Lubin Oct 9 at 18:38
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• I understand that in Z2, the only elements are 0 and 1. 0^2=0 and 1^2=1. Is this all I need to prove that if t is a root, so is t^2? Or am I missing something? Thanks! – Candace Shortt Oct 9 at 18:54
• You're missing something. The coefficients of your polynomial need to be in $\Bbb Z_2[x]$ but $t$ could be in some extension of that field. I suggest you edit your question to include the thought you expressed in your comment. That will help demonstrate that you are attempting to solve the problem yourself. – Robert Shore Oct 9 at 19:41
• @CandaceShortt hows it looking? Still need any help? – C. Brendel Oct 10 at 8:12

HINT:

The fact that $$t$$ is a root means that

$$f(t)=t^n+t^{n-1}+\ldots t^2+t+1=0.$$

Over the given field you have that

$$(a+b)^2=a^2+b^2$$

which is extendable for more terms, so

$$(a+b+c)^2=a^2+b^2+c^2$$

and so on.

What happens when you take

$$t^n+t^{n-1}+\ldots t^2+t+1=0$$

and raise both sides to the power of $$2$$?

Compare it with $$f(t^2)$$.

Hope this helps, otherwise ask in the comments

Given a polynomial $$f\in\mathbb{Z}_2[x]$$ it doesn't have to have a root in $$\mathbb{Z}_2$$. An example would be $$f=x^4+x+1$$ or more applicable to your case $$g=x^4+x^3+x^2+x+1$$. You can check that just by plugging $$0$$ and $$1$$ into the equations: $$f(0)=0^4+0+1=1\ne0\text{ and } f(1)=1^4+1+1=3=1\ne 0$$ aswell as $$g(0)=1\text{ and } g(1)=1$$ What now? There's a theorem in algebra that says that for any polynomial $$f$$ of degree $$n$$ over some field $$K$$ there is a bigger field $$L$$ that contains $$K$$ such that $$f$$ splits into exactly $$n$$ linear factors. The smallest such field is called to splitting field of $$f$$ over $$K$$. But splitting into linear factors means we have zeros in this extension field of $$K$$. The standard example is $$f=x^2+1$$ over the field $$\mathbb{R}$$ where $$f$$ does not have roots! The extension field where $$f$$ splits turns out to be $$\mathbb{C}$$, i.e. $$f=x^2+1=(x+i)(x-i)$$, so we have the roots $$i$$ and $$-i$$ in the extension $$\mathbb{C}$$ of $$f$$ over $$\mathbb{R}$$.

As it turns out every field $$L$$ (not only a splitting field of some polynomial) containing the field $$K$$ has the same characteristic as $$K$$, so $$\text{char}(L)=\text{char}(K)$$.

So now turning to your problem: Given the polynomial $$f=x^n+x^{n-1}+\dots+x+1$$ in $$\mathbb{Z}_2[x]$$ there is a field $$L$$ containing $$\mathbb{Z}_2$$ such that $$f$$ has exactly $$n$$-roots in $$L$$. Your exercise is probably to show that given a root $$t\in L$$ then $$t^2$$ is also such a root (If your exercise restricts the variable purely to $$\mathbb{Z}_2$$ then the exercise would be trivial, as $$t^2=t$$ for either $$0$$ or $$1$$. So lets assume otherwise). We find that: \begin{align} f(t^2)&=(t^2)^n+(t^2)^{n-1}+\dots+(t^2)+1\\&=(t^n)^2+(t^{n-1})^2+\dots+t^2+1^2 \end{align} Notice that in the last line I replaced $$1$$ with $$1^2$$. As hinted at in your exercise we have that $$(x+y)^2=x^2+y^2$$ in $$\mathbb{Z}_2$$. This holds in any field of characteristic $$2$$ (we have the same effect with $$(x+y)^p$$ for a field of characteristic $$p$$ - this is called the Frobenius homomorphism if you are intersted) and as we said any field extension of $$\mathbb{Z}_2$$ also has characterstic $$2$$ so especially our splitting field $$L$$ and we can use the neat fact of Frobenius: \begin{align} f(t^2)&=(t^n)^2+(t^{n-1})^2+\dots+t^2+1^2\\&=(t^n+t^{n-1}+\dots+t+1)^2\\&=(f(t))^2\\&=0^2\\&=0 \end{align} so $$t^2$$ is a root of $$f$$ aswell!