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Why is 2=0 in K for part a)? Also I don't understand what part b) is asking you to do - what does it mean by alpha = [X], so M=etc. Could someone please explain the question and the solutions to part b) and c).

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3 Answers 3

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The equation "$2 = 0$" really means that the residue class of $2$ is the residue class of $0$; that is, $2$ and $0$ are just (customary) shorthand for these residue classes. Your coefficient ring $\mathbb{Z}_{2}$ is the field of integers modulo $2$, which consists only of (the residue classes of) $0$ and $1$. These residue classes are, by definition, the sets $\{\ldots, -4, -2, 0, 2, 4, 6, \ldots\}$ and $\{\ldots, -3, -1, 1, 3, 5,\ldots\}$. In other words, modulo $2$, we have $2 = 0$. This is really just a notational shorthand for $\overline{2} = \overline{0}$, where (e.g.) $\overline{2}$ is the residue class of $2$.

Part (b) is asking you to consider the quotient ring $M = K[X]/(f)$, where $(f)$ is the ideal in the polynomial ring $K[X]$ generated by your polynomial $f$. Here the $\alpha = [X]$ seems to denote the coset $X + (f)$ in this quotient ring. This quotient ring is finite, and in fact consists of just those four elements: $0 = [0] = 0 + (f) = (f)$, $1 = [1] = 1+(f)$, $\alpha = [X] = X+(f)$ and $\alpha+1 = [X+1] = X+1+(f)$.

Now, in this quotient ring, we have (e.g.) $\alpha^2 = [X]^2 = [X^2] = [-1] = [1]$. That is why there is a "1" in the row and column labeled $\alpha$ of the multiplication table. The other entries are obtained similarly, by "reducing" the calculations using the relation $\alpha^2 = 1$ until you get something linear in $\alpha$.

In part (c), it seems you are meant to notice that if, in the polynomial ring $K[X]$, the polynomial $X^2 + 1$ divides $X^{2009} + 1$, then the image of the former would divide the image of the latter in the quotient ring $M$ of $K[X]$. By showing that this does not happen in $M$, via the given calculations, you've shown that $X^2 + 1$ doesn't divide $X^{2009} + 1$ in $K[X]$.

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You are working in modular 2. In modular $2$, $2=0$. This leads to only having 0 and 1 available since $3=2+1=0+1=1; 4=2+2=0+0=0$, etc...

In part b you need to multiply the only four elements of $Z_2/f$. Remembering part a; modulo $f$ means that our polynomial, $f$, is equivalently $0$ now. This leads to a relationship $x^2+1=0$ which implies $x^2+1+1=0+1$ so $x^2=1$. So any power of $x$ greater than 2 can be rewritten as either $1$ if the power is even or $x$ if the power is odd. Try working out the tables with this relationship. Your results will simplify into the answers shown.

In part c. we use our relationships that $x^2=1$ and $2=0$ to do some algebra.

Since $(x^2)^1$$^0$$^0$$^4$$*x$ and $x^2$=1, $(x^2)^1$$^0$$^0$$^4$$*x$$=1^1$$^0$$^0$$^4$*x$=x$

So $x^2$$^0$$^0$$^9$$-1=x+1$; we now consider $x^2+1=1+1$(by our relationship) which $=0$. And 0 does not divide $x+1$

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For (a), the ring $\mathbb{Z}_2$ is the integers modulo $2$. That means we identify an integer with its remainder upon division by 2. Since $2$ is divisible by 2, it leaves remainder $0$, so $2\equiv 0\pmod{2}$, or $2=0$.

For (b), $\alpha=[X]$ is the image of $X$ in the quotient $K[X]/(f)$. This means we identify $\alpha^2+1=0$, or $\alpha^2=-1=1$, the last equality follows since we are working over $\mathbb{Z}_2$. So since every term of degree $2$ or higher may be reduced, so all polynomials in $M$ are of degree at most $1$, and since the only possible coefficients are $0$ or $1$, this gives you the four polynomials in $M$. The multiplication table follows by multiplying the polynomials as you would normally, and then reducing modulo $f$.

For (c), if $X^2+1$ were to divide $X^{2009}-1$, then $\alpha^{2009}-1=0$ in $M$. This follows for if $X^{2009}-1=g(X)(X^2+1)$ for some $g(X)$, then $$ \alpha^{2009}-1=g(\alpha)(0)=0 $$ after projecting the polynomials into $M$. But as the solution shows, $\alpha^{2009}-1$ is not $0$ in $M$, so $X^{2009}-1$ is not divisible by $X^2+1$ in $K[X]$.

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