I know that since $\Bbb Z$ is commutative and unital $\langle 7\rangle$ is just the polynomials whose coefficients are all in $7\Bbb Z$.  I know that since $\Bbb Z[x]$ is commutative that in order to show $\langle 7\rangle$ is a prime ideal show: $ab \in \langle 7\rangle$ implies $a \in \langle 7\rangle$ or $b \in \langle 7\rangle$ for all a,b in $\Bbb Z[x]$.  I have calculated what form ab is using Cauchy product. I now have stated each coefficient is in $7\Bbb Z$ but Im finding it hard to show how this requires either a or b lies in $\langle 7\rangle$
22880
Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable over $\Bbb Z$.  My question: Is every prime ideal of $\mathbb{Z}[X]$ one of following types? If yes, how would you prove this?  $(0)$.  $(f(X))$, where $f(X)$ is an irreducible polynomial.  $(p)$, where $p$ is a prime number.  $(p, f(X))$, where $p$ is a prime number and $f(X)$ is an irreducible polynomial modulo $p$.
67264
Quoting " Let $\phi : \Bbb Z \rightarrow \Bbb Z$ be given by $\phi(n) = 7n$. Prove that $\phi$ is a group homomorphism. Find the kernel, and the image of $\phi$."   My understanding:  Part 1:   Given $n,m \in \Bbb Z$, let's check that $\phi(n+m) =\phi(n)+\phi(m)$  $$\phi(n+m) = 7(n+m) = 7n + 7m = \phi(n)+\phi(m)$$  Therefore as the group operation in $\phi$ is preserved, $\phi$ is a group homomorphism.  Part 2 (revised):   $\phi$ is one-to-one as:  $\space \space \phi (n) = \phi(m) \Rightarrow 7n =7m \Rightarrow n=m  $. We also know that $\phi(e)=e$ by the property of homomorphism. Therefore, no other object than $0$ in the domain can map to $0$ in the codomain. It follows that: $$\ker(\phi)=\{ x \in \Bbb Z : \phi(x)=e\}=\{e\}$$   Part 3 (revised):  I claim that the only possible image-elements of $\phi$ are the multiples of $7$ in $\mathbb Z$ denoted $7\mathbb Z$.   To prove this I have to show that Im$(\phi) \subset 7 \Bbb Z$ and that $ 7 \Bbb Z \subset $ Im$(\phi)$.  Proving Im$(\phi) \subset 7 \Bbb Z$ :      $\forall y \in $ Im$(\phi)$ such that $  y =\phi(x)$ where $x \in 7 \Bbb Z$. It follows that $y \in 7 \Bbb Z$, therefore Im$(\phi) \subset 7 \Bbb Z$   Proving $ 7 \Bbb Z \subset $ Im$(\phi)$:   $\forall y \in  7\Bbb Z$ such that $  y =7x$ where $x \in \phi(x)$. It follows that $y \in \phi(x)$, therefore $ 7 \Bbb Z \subset $ Im$(\phi)$.   Any input to my understanding is much appreciated.
40062
Let $I = \langle 2\rangle$. Prove $I[x]$ is not a maximal ideal of $\mathbb Z[x]$ even though $I$ is a maximal ideal of $\mathbb Z$.   My professor mentioned that I should try adding something to it to prove that $I[x]$ is not the maximal ideal. I want to assume he meant to add $ \langle 2 \rangle$ to $I[x]$?
2257
Suppose that we are dealing with the real field $\mathbb{R}$ and we define $T:\mathbb R^2\to\mathbb R^2$ by $$T(u,v):=(-v,u).$$ It is trivial to see that $\langle(u,v),T(u,v)\rangle = 0$ for all $u, v \in \mathbb{R}$, where $\langle \, \cdot \, , \, \cdot \, \rangle$ is the standard inner product.  A question in Hoffmann-Kunze asks us to define all possible inner products $[ \, \cdot \, , \, \cdot \, ]$ such that $[\alpha, T(\alpha)]=0$ for all $\alpha\in\mathbb R^2$ and $T$ is as above. How does one find all such inner products?
38567
The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.   I don't know how to consider it. Any suggestions?
91590
Find $|\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle|$ and determine if its cyclic.   $\mathbb{Z} \oplus \mathbb{Z} = \{(1,1), (1, 2),... (2, 1),...(3,1),...\}$  $\langle(3,6)\rangle = \{(0,0),(3,6),(6,12),...\}$  $\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle = \{(a,b) + \langle(3,6)\rangle : a,b \in \mathbb{Z}\}$  How would I go by finding the order of the quotient group? Would it be infinite? Since both sets that make up the quotient group are infinite, does that mean that the order of the quotient group is also infinite? What are some things I'm not noticing that could be helpful?
15412
I was having some trouble finding an explicit isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$.  $\textbf{What I have noticed is}:$   7 is a prime element in $\mathbb{Z}[i]$ so $(7)$ is a maximal ideal in $\mathbb{Z}[i]$ and $\mathbb{Z}[i]/(7)$ is  a field.  7 is also a prime element in $\mathbb{Z}[\sqrt{-2}]$   $\textbf{What I have been trying to do is this}$   Find a surjective homomorphism between $\mathbb{Z}[i]/(7) \rightarrow \mathbb{Z}[\sqrt{-2}]/(7)$. Since $\mathbb{Z}[i]/(7)$ was a field, the kernel of this homomorphism will be either the whole ring or just $0$. In the latter case it would be a isomorphism.    I am having trouble finding this surjective homomorphism:  I have noticed that $\bar{{i}}^{2}=-1$ so the image of $\bar{{i}}$ must be sent to something whose square is $-1$. Any help would be appreciated. I may be missing some obvious insight.
69770
I'm asked to prove that the quotient ring $\mathbb{Z}[x]/\langle x, 2\rangle$ is isomorphic to $\mathbb{Z}_2$.  The first thing that came to my mind was to apply the Theorem of Homomorphism of Rings, but to do so I need to construct a homomorphism $\phi: \mathbb{Z} \to \mathbb{Z}_2$, such that $\mathrm{Ker}\phi = \langle x, 2\rangle$ ($\langle x, 2\rangle$  is an ideal of $\mathbb{Z}[x]$). But     What elements are in $\mathbb{Z}[x]/\langle x, 2\rangle$ in the first   place?   I know that $\langle x, 2\rangle$ of all the polynomials with even constant term, but I can't figure out how the elements of $\mathbb{Z}[x]/\langle x, 2\rangle$ look like.  Any help would be appreciated.
77054
I am trying to show the isomorphism  $$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$  I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
99583
Here is a problem I face practicing the theory of rings:     Define $\phi : \mathbb{Z}[t] \to \mathbb{Q}$, a ring homomorphism (it does map $1$ to $1$). I'm trying to show that if $\phi(t)=\frac{u}{v}$ (in lower terms) then we have $\frac{m}{n}\in\operatorname{Im}(\phi)$ (in lower terms again) if and only if every prime factor of $n$ divides $v$ (in lower terms).   Then I'd like to deduce from this that there does not exist any ideal $I$ of $\mathbb{Z}[t]$ such that $\mathbb{Z}[t]/I$ is isomorphic to $\mathbb{Q}$.  I have:  $$\phi(\sum_{i=0}^{N} a_i t^i)=\sum_{i=0}^{N} a_i\phi(t)^i.$$  If $\frac{m}{n} \in\operatorname{Im}(\phi)$ then there are $N \in \mathbb{N}$ and $a_i\in\mathbb{Z}$ ($i=0,1,...,N$) such that $mv^N=n\sum_{i=0}^{N} a_iu^iv^{N-i}$. Hence one of the implications is quite straightforward as if $p$ prime divides $n$ then it divides the left-hand-side and as it does not divide $m$ it divides $v^N$ and so $v$. But for some reason I can't get to the other way around. I've tried to 'find' the coefficients $a_i$ and show they were integers for some $N$ but I couldn't.  Also, I'm not sure at all how to make progress on the deduction...  Could you hint me please? I'd be very grateful, this is getting a little frustrating...  Thank you!
3951
A very basic ring theory question, which I am not able to solve. How does one show that    $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$. Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.   My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.
95738
$\mathcal P_4(\mathbb R)$ is the set of polynomials with degree at most $4$ with real coefficients.  $U=\{p \in \mathcal P_4(\mathbb R) \mid p(6)=0\}$.  Would $\langle(x-6),(x-6)^2, (x-6)^3, (x-6)^4\rangle$ form a basis?  Suppose $a(x-6)+b(x-6)^2+c(x-6)^3+d(x-6)^4=0$. Then we know $a=b=c=d=0$ since the RHS has no $x, x^2, x^3, x^4$ terms. So the set of vectors is linearly independent.  Also, since $\dim U &lt; 5$, we cannot extend a basis of $U$ beyond length $4$. Hence $\dim U=4$.  So, $\langle(x-6),(x-6)^2, (x-6)^3, (x-6)^4\rangle$ is a basis of $U$.
50159
This problem goes as follows:     Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions $\mathbb{Q}(t)$.    The first part is straightforward, $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, and  hence a prime ideal, as $\mathbb[Q][x,y]$ is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain.   How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how $x^2+y^2-1$ is going to help me make a choice.
20389
I'm trying to prove that Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).  I know that Aut $\mathbb Z_p$ has $p-1$ elements because $\mathbb Z_p$ has $p-1$ possiblities of generators, so intuitively I see Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$, but I couldn't prove it formally.  I'm trying to build an isomorphic function, but I don't how to do it.  Am I in the right way? Maybe I'm forgetting some trick or something.  Thanks
46016
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:     Let $R$ be a unitary commutative ring, and $I$ be an ideal, and $N$ is an $R$-module. I'll now show that $$I \otimes_R N \cong IN\,.$$   First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$. So then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$  But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?
113988
Give $\ker(\phi)$  for $\phi: \mathbb{Z} \to \mathbb{Z_7}$ such that $\phi(1) = 4$  So the generator of $\mathbb{Z}$ maps to $4$ and $\langle 4 \rangle = \{0,1,2,3,4,5,6\}$ so it has order seven.  so $\ker(\phi) = 7\mathbb{Z}$ because every seventh element will map to the identity under the given map. Is that the correct reasoning?
62747
Let $\mathbb Z$ be the ring of integers. The question asks to show  that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.