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I have this abstact algebra problem: considering $ A=\frac{\mathbb Z_2[X] }{(X^2+X-1)}$ and I have to prove that it is a domain. In order to be a domain I thought that it is supposed to be a commutative ring, to have a unit and to have no zero-divisors but i don't know how to start considering it is a ring of polynomials . thanks for the help !

Bill Dubuque
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andrea98
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2 Answers2

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You know that $\mathbb{Z}/2\mathbb{Z}$ is an integral domain. Now prove that (or use a fact already known to you) that $\mathbb{Z}/2\mathbb{Z}[x]$ is an integral domain. Quotients of commutative rings are commutative. So your ring is commutative. You only need to show that has no zero divisors.

Recall (or prove if you have not seen this) that if $R$ is a commutative ring with identity and $I$ is an ideal of $R$, that $R/I$ is an integral domain if and only if $I$ is prime. So apply this to $R= \mathbb{Z}/2\mathbb{Z}[x]$ and $I=(x^2+x-1)$, i.e. show that this $I$ is a prime ideal. This can be as simple as showing it does not factor.

However because it seems that you do not recognize the elements in your ring, it may be useful to write out all the elements and their multiplications (in a table) to prove this and get an idea of how a quotient ring works. Note that your ring is the set of all such $a_1 x+a_0$, where $a_1, a_0 \in \mathbb{Z}/2\mathbb{Z}$. This is because the elements of your ring are the possible remainders after long division by $x^2+x-1$ over the ring $\mathbb{Z}/2\mathbb{Z}$. So perhaps remind yourself of long division of polynomials in $\mathbb{R}[x]$, then try your examples using $\mathbb{Z}/2\mathbb{Z}$ (i.e. binary) instead of $\mathbb{R}$, then convince yourself that this set is correct. Then write out the additions/multiplications of these elements. This may be a good way for an initial approach to get an idea of quotient rings for polynomials, but ultimately you want to understand the 'theory' approach too. In fact, once you do, see what property of $x^2+x-1$ you needed. Can you write down other polynomials where this will still be true? How about polynomials where the result will not be an integral domain? Can you even find explicit zero divisors in these cases? Can you see how these zero divisors relate to the polynomials?

mathematics2x2life
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  • i can't understand this part "This is because the elements of your ring are the possible remainders after long division by x2+x−1 over the ring Z/2Z. " – andrea98 May 12 '20 at 07:43
  • @AndreaZannoni This is the definition of a quotient ring at work. We know in a quotient ring that two elements $r,r'$ are equivalent if $r+I= r'+I$, i.e. $r-r' + I= I$. But because you can long divide, you can take any polynomial $p(x)$, long divide by your polynomial, and write $p(x)= q(x) \cdot (x^2+x-1) + r(x)$, where $r(x)$ has degree less than $x^2+x-1$. But mod $x^2+x-1$, you have $p(x) \equiv r(x)$ because $(x^2+x-1) q(x)\equiv 0 \cdot q(x) \equiv 0$ in the quotient ring. – mathematics2x2life May 12 '20 at 07:47
  • okay thank you!! – andrea98 May 12 '20 at 07:53
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$\mathbb Z_2$ is a field. Hence $\mathbb Z _2[x]$ is a PID. Now observe that $x^2+x-1$ is a polynomial of degree 2 and it is reducible in $\mathbb Z _2[x]$ iff it has a root in $\mathbb Z_2$.

But $x^2+x-1$ doesn't have a root in $\mathbb Z_2$.

Hence $x^2+x-1$ is irreducible in $\mathbb Z _2[x]$.

So $(x^2+x-1)$ is a maximal ideal hence $\frac{\mathbb Z _2[x]}{(x^2+x-1)}$. ( Quotient by a maximal ideal for a commutative ring with identity is a field)

Noob mathematician
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