3
$\begingroup$

I need help understanding the following isomorphisms:

$$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$

In general, I am also wondering whether the following are true, and if so briefly why:

  • $\mathbb Z[\sqrt{D}] \cong \mathbb Z[x]/(x^2 - D)$ for all squarefree integers $D$
  • $(R/I)/J \cong R/(I +J)$, ie why do we have things like $\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+5) $ where $\mathbb Z[x] / (2) \cong \mathbb Z_2[x]$?

Also for example for something like the following:

$$\,\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+1)$$

Is it true because of the third isomorphism theorem? I'm guessing we set $R = \mathbb Z[x]$, $I = (2)$, and $J = (x^2 + 5)$, but please expand more on each step of the reduction.

  • 0
    $\phi(a+b \sqrt{D}) = a+bx$ is an isomorphism $\mathbb Z[\sqrt{D}] \to\mathbb Z[x]/(x^2 - D)$. Also in arbitrary integral domain, $\sqrt{D}$ is really a root of $x^2-D$, nothing more, so $R[\sqrt{D}] \cong R[x]/(x^2 - D)$ by definition. And show that for some $a \in R$ with image $\tilde{a} \in R/I$ then $(R/I)/(\tilde{a}) = R/(I,a)$ – 2017-02-08
  • 0
    Strongly related: http://math.stackexchange.com/questions/1871692 – 2017-02-08

3 Answers 3

2

Yes all this results from the 3rd isomorphism theorem. Here are some details: \begin{align}\mathbf Z[\sqrt{-5}]/(2,1+\sqrt{-5})&\simeq\mathbf Z[X]/(X^2+5)/(2,1+X)\cdot Z[X]/(X^2+5)\\&\simeq\mathbf Z[X]/(X^2+5)/(2,1+X, X^2+5)/(X^2+5)\end{align} For the second isomorphism: \begin{align}\mathbf Z[X]/(2,X^2+5)&\simeq\mathbf Z[X]/2\mathbf Z[X]/(2,X^2+5)\mathbf Z[X]/2\mathbf Z[X]\\ &\simeq\mathbf Z/2\mathbf Z[X]/(X^2+1)\mathbf Z/2\mathbf Z[X] \end{align}

  • 0
    Why do we have $IR/J = (I + J)/J$? – 2017-02-08
  • 0
    Because it is the set of all $i+J$ ($i\in I$), and for any $j\in J$, we have $i+J=i+j+J$. – 2017-02-08
  • 0
    Okay, I almost understand everything, but could you explain why when we go from $Z[\sqrt{-5}]$ to $Z[x]/(x^2 + 5)$, the ideal $(2, 1 + \sqrt{-5})$ goes to $(2, 1 + x)\cdot Z[x]/(x^2 + 5)$? – 2017-02-08
  • 0
    $1+\sqrt{-5}$ is $1+x$ with the initial identification, so the ideal $(2,1+\sqrt{-5})\mathbf Z[\sqrt{-5}]$ is identified with $(2,1+x)\mathbf Z[x]/(x^2+5)$. Isn't it natural? – 2017-02-08
  • 0
    Dear, Bernard! Sorry, but what do you mean by $(2,1+x)\cdot Z[x]/(x^2+5)$? – 2018-12-08
  • 1
    The ideal generated in the quotient ring by the classes of $2$ and $1+x$. – 2018-12-08
  • 0
    Is it the same as $(2+J,1+x+J)$ where $J=(x^2+5)$, right? – 2018-12-08
  • 0
    More precisely, $\bigl((2)+J, (1+x)+J\bigr)$ (sums of ideals, not congruence classes of elements). – 2018-12-08
  • 0
    Things became quite unclear. What do you mean here by $(2)$? What ideal? – 2018-12-08
  • 0
    Since $Z[\sqrt{-5}]\cong Z[x]/(x^2+5)$ and we need to know what ideal corresponds to $(2,1+\sqrt{-5})$. Element $2$ corresponds to $2+(x^2+5)$ and $1+\sqrt{-5}$ to $1+x+J$. Hence the corresponding ideal in $Z[x]/(x^2+5)$ is $(2+(x^2+5),1+x+(x^2+5))$. What's wrong with my argument? – 2018-12-08
  • 0
    You have to consider the classes of all elements in the ideal generated by $2$ and $1+x$ – otherwise you obtain $2+J +1+x+J=3+x+J$ – a class, not an ideal in the quotient. – 2018-12-08
  • 0
    I cannot understand what's wrong with my reasoning in the previous comment? The $(2+(x^2+5),1+x+(x^2+5))$ is the ideal in quotient-ring generated by its elements $2+(x^2+5)$ and $1+x+(x^2+5)$. And also I didn't understand your example. – 2018-12-08
  • 0
    The problem is not with the reasoning, but with the notations: just take general notations – you have a quotient ring $R/J$ and two ideals $I_1, I_2$ in $R$. They generate in $R/J$ the ideal $$(I_1+I_2\cdot R/J=(I_1+J+I_2+J)/J=(I_1+I_2+J)/J.$$ – 2018-12-08
  • 0
    Sorry for that I have some difficulties. In our example $R/J$ is $Z[x]/(x^2+5)$, right? but what is $I_1$ and $I_2$? – 2018-12-08
  • 0
    Don't be sorry, we're here to help. They're the ideal $(2)$. and the ideal $(1+x)$ – 2018-12-08
  • 0
    If $I_1=(2), I_2=(1+x)$ in $\mathbb{Z}[x]$ then they generate an ideal $((2)+(1+x)+(x^2+5))/(x^2+5)=(2,1+x,x^2+5)/(x^2+5)$ in $\mathbb{Z}[x]/(x^2+5)$. Am I right that this ideal corresponds to an ideal $(2,1+\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$? If yes, it is not obvious to me why it's true. But i have the following reasoning: suppose $\phi:\mathbb{Z}[\sqrt{-5}]\to \mathbb{Z}[x]/(x^2+5)$ is desired isomorphism then the mapping $\phi$ restricted to ideal $(2,1+\sqrt{-5})$ defined by $\phi: 2(a+b\sqrt{-5})+(1+\sqrt{-5})(c+d\sqrt{-5})\mapsto 2(a+bx)+(1+x)(c+dx)+J$, – 2018-12-08
  • 0
    where $J=(x^2+5)$ - ideal. Am I correct? – 2018-12-08
  • 0
    Yes, ${}$exactly. – 2018-12-08
  • 0
    I guess, that finally I've got you. By $(2,1+x,x^2+5)/(x^2+5)$ we mean the following set $\{2f+(1+x)g+(x^2+5)h+\langle x^2+5\rangle: f,g,h\in Z[x] \}$. In order to not to confuse you and myself by $\langle x^2+5\rangle$ I denoted the principal ideal generated by $x^2+5$. – 2018-12-08
  • 0
    Dear, Bernard! I am very grateful to you for your great help and patience because I've asked some stupid questions. But it was important to me understand these things. Thank you very much! :) – 2018-12-08
  • 0
    Yes you got me. Sorry for not having been clearer. Sometimes it's not easy to guess what people asking a question are familiar with or not. – 2018-12-08
1

For example:

$$\phi:\Bbb Z[x]\to\Bbb Z_2[x]\;,\;\;\phi(p(x)):=p(x)\pmod 2$$

is a rings epimorphism (reduction modulo $\;2\;$ of all the coefficients in the polynomial), and its kernel is clearly $\;(2)=2\Bbb Z[x]=\;$ the polynomials whose coefficients are even numbers, so by the first isomorphism theorem you have $$\Bbb Z[x]/(2)\cong\Bbb Z_2[x]$$

Then observe that $\;x^2+5=x^2+1=(x+1)^2\pmod 2\;$ and you get the last result.

For the first result I think the following is the easier approach: since using the FIT again, as above, we get

$$\Bbb Z[x]/(x^2+5)\cong\Bbb Z[\sqrt{-5}]$$

the trick used in the first line is to identify $\;\sqrt{-5}\approx x\;$ in the quotient, so

$$\Bbb Z[\sqrt{-5}]/(2,\,1+\sqrt{-5})\cong\left(\Bbb Z[x]/(x^2+5)\right)/(2,\,1+x)\cong$$

$$\cong\Bbb Z[x]/(x^2+5,\,2,\,1+x)\cong\Bbb Z[x]/(2,1+x)$$

the last step following by the fact that $\;(2,\,1+x)\;$ is a maximal ideal in $\;\Bbb Z[x]\;$ and thus $\;\Bbb Z[x]/(x^2+5,\,2,\,1+x)=\Bbb Z[x]/(2,\,1+x)\;$ , since for example

$$(x-1)(x+1)+2\cdot3=x^2+5\in(2,\,1+x)$$

  • 0
    Would you mind explicitly answering the questions in the bullet points? Thanks! – 2017-02-08
  • 0
    Also, how can we see that $(2, 1+x)$ is maximal? – 2017-02-08
  • 0
    @i The first bullet is **exactly** the same as done with the example with $\;\sqrt{-5}\;$ and the FIT. Try to construct explicitly a homomorphism and etc. The second bullet makes sense if $\;J\le R/I\iff I\le J\le R\;$, so that's simply the Third Is. Th. As for why $\;(2,1+x)\;$ is maximal: in fact, it is because $\;2\;$ is a prime and $\;1+x\;$ is irreducible in $\;\Bbb Z/2\Bbb Z[x]=\Bbb Z_2[x]\;$. You may want to check here: https://www.ma.utexas.edu/users/voloch/Homework/zx.pdf – 2017-02-08
0

$\mathbb Z[\sqrt{-5}]\cong \mathbb Z[x]/(x^2+5)\Rightarrow \mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5})\cong \mathbb Z[x]/(x^2+5,2,1+x) \cong \mathbb Z_2[x]/(x^2+5,1+x)= \mathbb Z_2[x]/(x^2+1,x+1)$

we have $x^2+1=(x+1)(x+1)$ in $\mathbb Z_2$, so $x^2+1\in (x+1)$, ((or $(x^2+1,x+1)=(x^2+1(mod (x+1), (x+1))=((-1)^2+1,x+1)=(0,x+1)=(x+1)$))

and $\mathbb Z_2[x]/(x+1)\cong \mathbb Z_2[-1]=\mathbb Z_2[1]=\mathbb Z_2 $, so $\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5})\cong\mathbb Z_2 $