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I have the following question :

$K=\frac{\mathbb{Z}_5[x]}{(f)}$ is a field with 125 elements.

$\lambda=x+(f)\in K$

$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,$ and $\lambda^{625}$ express $\lambda^{625}\in K$ in the following form $a_0+a_1 \lambda+a_2\lambda ^2$ while $a_0,a_1,a_2\in \mathbb{Z}_5$

I did manage to find $\beta$ which is $0$ since there's a theorem that say $q^n=125$ (125-number of elements) so the max power should be $3$, Yet I don't know how to find $\alpha$ and $\lambda^{625}$

Any help will be appreciated.

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    @peterag Yes, sorry edited – 2017-01-30
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    Hint(s): The choices for $\alpha^2$ are $1$ or $4=-1$. Now, for $K$ to be field, $f$ must be irreducible over the ground field. Since $f$ is of degree $3$, that means that $f$ cannot have a root (over the ground field). Hint for the 2nd question - long division... divide $x^{625}$ by $f$ [of if you prefer, by a monic version of $f$]. OK? – 2017-01-31
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    @peterag The first question I understand how you found $\alpha$ but the second question is still not clear divide $\lambda^{625}=(x+f)^{625}$ in f? – 2017-01-31
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    First of all, just to be clear, $\lambda = x + (f)$, i.e., with the parentheses around $f$. [in your comment, you dropped the parentheses.] That is, $\lambda$ is the image of $x$ under the ring homomorphism (quotient map) $$k[x]\rightarrow K= k[x]/ (f),$$ where $k$ is the base field. In particular, $f(x)$ goes to $0$ under the map $$x\mapsto \lambda,$$ or equivalently, $f(\lambda) = 0$, when calculated in $K$. (cont) – 2017-01-31
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    (cont) Now: if one does (long) division (divide by $f$) in $k[x]$, you'll end up with an expression $$ x^{625} = q(x) f(x) + r(x),$$ where the degree of $r$ is less than the degree of $f$. So, [as $f(\lambda)=0$], one has $\lambda^{625} = r(\lambda)$ in $K$. OK? – 2017-01-31
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    Or if you prefer, for what it's worth, you can also divide out by $-f$... Also, again for what it's worth, since $y^{125} =y $ in $K$ (for all $y\in K$), and thus $\lambda^{625} = \lambda^{125\cdot 5} =\lambda^5$, you can instead calculate the remainder of $x^5$ after division by $f$; it should give you the same answer. – 2017-01-31
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    I understand that $\lambda^{625}=\lambda^5$, so we get that $\lambda^{625}=\lambda^5=(x+(f))^5=(3+3x+4x^3)^5$ but to divide $(3+3x+4x^3)^5$ by $f$? why? I think I misunderstood what you meant. – 2017-01-31
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52799/discussion-between-peter-a-g-and-javapg). – 2017-01-31
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    you are a bit confused about something - please let's chat... see previous comment. I have a few mins for it if you do – 2017-01-31
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    thinking back to the link you sent - just a quick word: I wrote 'basically' but by long division, I meant the 'non'-denominator version . For instance for $\mathbb Z/(5)$, you don't use $13/5 = 2 + 3/5$; rather $13 = 2\cdot 5 + 3$, as we want the remainder $3$. Ditto $p(x) = q(x) \cdot f(x) + r(x)$, rather than $p(x)/f(x) = q(x) + r(x)/f(x)$ for $k[x]/(f(x))$, Good luck! – 2017-01-31

2 Answers 2

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$\lambda\ne0$ implies $\lambda^{124}=1$ by Lagrange's theorem applied to the multiplicative group of $K$.

Then $\lambda^{625}=\lambda^{5}$, which makes it easier, using that $3+2\lambda+\alpha^2\lambda^3+\beta \lambda^4=0$.

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We have $\frac{\mathbb Z_5[x]}{(f)}\cong \mathbb Z_5(a)$ then number of the element in $\mathbb Z_5(a)=125=5^3$ then $[\mathbb Z_5(a):\mathbb Z_5]=3$ then $f(x)$ is a polynomial of degree $3$ and donot have roots in $\mathbb Z_5$, then $\beta=0$.

we have $\alpha\in \mathbb Z_5 $,then $\alpha^2=0,1,4 $

if $\alpha^2=0$ then $f(x)=3+2x$,then $deg(f)<3$ contradiction

if $\alpha^2=1$ then $f(x)=3+2x+x^3 $, then $2$ root of $f(x)$ contradiction

if $\alpha^2=4$ then $f(x)=3+2x+4x^3$ and donnot have a root in $\mathbb Z_5$

Let $a$ root of $f(x)$ in $ \mathbb Z_5(a)$ then $\lambda=x+(f)$ root in $\mathbb Z_5[x]/(f)$, for calculate $\lambda^{625}$ we must to calculate $a^{625}$

we have $4a^3+2a+3=0\Rightarrow a^3=2a+3\Rightarrow a^4=2a^2+3a \Rightarrow a^5=3a^2+4a+1\Rightarrow a^6=4a^2+2a+4 $, so $a^{10}=3a+3\Rightarrow a^{20}=4a^2+3a+4\Rightarrow a^{25}=2a^2+4\Rightarrow a^{50}=4a^2+2a+1 $, so $ a^{50}=a^6+2 \Rightarrow a^{100}=(a^6+2)^2=a^{12}+4a^6+4=4a^2+4a+4\Rightarrow a^{100}=a^6+2a $,so $a^{200}=(a^6+a)^2=4a+2 \Rightarrow a^{400}=a^2+a+4\Rightarrow a^{600}=a^2+a$.

Therefor $a^{625}=a^{600}a^{25}=3a^2+4a+1=a^5$

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    Hi Mustafa - please see my comment to the original question! For $K$ to be a field, $f$ must be irreducible. Hence it CANNOT have a root in the ground field. So I would say that your calculation actually implies the opposite of what I understand you to be saying: i.e, $\alpha^2$ must NOT be equal to $1$. Therefore $\alpha^2=4=-1$ OK? – 2017-01-31
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    @ peter a g - Thank you for comment, I edit my answer. – 2017-01-31
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    I edited your layout, especially because there was a typo that caused the last exponent $5$ of $a^5$ not to show up. I hope you don't mind! (I added a few 'so's because the long math mode was looking odd in my browser). – 2017-01-31
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    no, I don't mind, Thank you very much for last comment and edit, thanks again – 2017-01-31
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    And don't you mean $a^3 = 3 + 2a$, and thus with the subsequent changes? – 2017-01-31
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    ok ,soory, I don't notice, I will edit my answer – 2017-01-31