ok, so your ring is isomorphic to $\mathbb{Z}[X] / (X^2 + 1, 2X+2)$?, and $\mathbb{Z}[X]/(2X+2)$ is isomorphic to $\mathbb{Z}/(2) [X] \oplus \mathbb{Z}[X]/(X+1)$, the right summand is isomorphic to $\mathbb{Z}$, and now quotienting by $X^2+1$ gives us makes the left summand $\{ai + b : (a,b) \in \{0,1 \} \}$ with multiplication mod $2$, so four elements, and on the right you get $\mathbb{Z}_{2}$
so overall your thing should have $8$ elements
technically i dont think this solution requires any knowledge about stuff you haven't come across
besides that quotienting by $(a,b)$ means you can do so 'consecutively'
so for the first isomorphism, the reason its iso to your ring is $\mathbb{Z}[i]$ is the same as $\mathbb{Z}[X] / (X^2+1)$ which you can basically see by long division, and then by quotienting out $2X+2$ its akin to quotienting $\mathbb{Z}[i]$ by $2i+2$
and for the direct sum decomposition, its just the chinese remainder theorem
but idk if you have learned that
oh crap, this is probably wrong again, i dont think I can use the CRT here
ok, so we cant use CRT, then maybe we can check $\mathbb{Z}[X]/(2X+2) \rightarrow \mathbb{Z}/(2)[X] \oplus \mathbb{Z}[X]/(X+1)$ is anyway an isomorphism, this boils down to showing if an integer polynomial is divisible by both $2$ and $X+1$, that it is divisible by $2X+2$, i think that holds because if $P = (X+1)Q$, then mod $2$, $P mod 2 = (X+1)(Q mod 2) = 0$ in $\mathbb{Z}/(2) [X]$ which is an integral domain, so $Q mod 2 = 0$
sorry - ignore what im saying, im clearly way too rusty lol, the CRT map is always injective...
Since $4$ is in $I$, it follows that $4+I=0+I$.
Dividing $a$ and $b$ by $4$ by Euclid's lemma, one gets the remainders $0\le a'<4$ and $0\le b'\lt 4$ and $a',b'$ can't be both zero because $a+ib$ is not in $I$. So the following holds:
$a+ib+I=a'+ib'+I$
Possible values for $(a',b')$ are $15$. Now,...
there is only one minor problem with the answer. That's fixable and doesn't affect much the solution. I'll do it some other time.
so i think this is the same as asking you to find all maximal ideals of K[x] containing (x-1) and (x-2), which are only (x-1),(x-2) since K is a field? Then the corresponding maximal ideals of R/I should be these ideals images
@lulu For your comment on the meta post, I found the formula inside of Ramanujan's collected papers, it was the paper on Bernoulli numbers. He mentions the formula from Edwards Differential calculus.
@LukasHeger I did this way. s(\pi \times \lambda) = s(\pi) s(\lambda) where s is sign of permutation. Because identity permutation is even. If we multiply odd number of permutations we will get odd sign so contradiction. Is this right?
How do I prove that for $0<t<1$ and $\varepsilon > 0$ there is a homeomorphism $h_n$ of $[-1, 1]^n\times [-1, 1]$ such that $h\restriction [-1, 1]^n\times [-1, t]$ is the identity and $\text{diam}(h([-1, 1]^n\times \{1\})<\varepsilon$
For $n = 1$ I know such exists. How do I prove this by induction
If I have a matrix $A = \begin{bmatrix} 0 & b\\ -b & 0 \end{bmatrix}$ can I not do a change of basis (reverse the basis vectors) so that I get $A = P^{-1} \begin{bmatrix} b & 0\\ 0 & -b \end{bmatrix} P$
A separate question though, when drawing the phase portrait for the system $x' = Ax$, why do we calculate the Jordan decomposition $J = T^{-1}AT$ and plot the phase portrait for $e^{Jt}$, does $e^{At}$ follow the same behaviour?
Since $Te^{Jt}\vec{x} = e^{At}T\vec{x}$ would it not be better to plot the phase portrait of $Te^{Jt}$
Yeah, I guess I'm trying to get my head around it, what if I had stability along the $x$-axis, but my change of coordinates is a swap of axes, do I still have stability along the $x$-axis?
Problem $1$: minimum 2-edge connected subgraph
We are given a $2$-edge connected undirected graph $G(V,E)$, and we are asked to find a $\textit{spanning}$ subgraph $H$, with minimum number of edges, such that $H$ is $2
$-edge connected.
There is a result by Jothi et. al. that there exists a $\f...
Hi everyone. I was reading an article on PlanetMath and I found a surprising theorem: https://planetmath.org/recursivelyaxiomatizabletheory
Basically, it says that all "recursively axiomatizable theories are decidable." Is this really true? I see how recursively axiomatizable theories are semidecidable because you can enumerate through all proofs, but I don't see why all recursively axiomatizable theories need to be decidable.
Problem $1$: minimum 2-edge connected subgraph
We are given a $2$-edge connected undirected graph $G(V,E)$, and we are asked to find a $\textit{spanning}$ subgraph $H$, with minimum number of edges, such that $H$ is $2
$-edge connected.
There is a result by Jothi et. al. that there exists a $\f...
