Linked Questions

1
vote
4answers
497 views

Maximal ideal/ Prime ideal [duplicate]

Show any maximal ideal is a prime ideal. I am not sure how to approach this, i know i need to show that for any 'ab' in the maximal ideal, a is in the max ideal or b is in the max ideal.
5
votes
9answers
922 views

If a prime $p\mid ab$, then $p\mid a$ or $p\mid b$

If a prime number $p$ is a divisor of a product $ab$, $p$ has to be a divisor of $b$ or $a$. How can I demonstrate this theorem? I demonstrated this theorem on one way using Bezout's theorem in an ...
5
votes
7answers
22k views

If $\gcd(a,b)= 1$ and $a$ divides $bc$ then $a$ divides $c\ $ [Euclid's Lemma]

Well I thought this is obvious. since $\gcd(a,b)=1$, then we have that $a$ does not divide $b$ AND $a$ divides $bc$. this implies that $a$ divides $c$. done. but apparently this is wrong. help ...
7
votes
2answers
8k views

Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$.

Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.
8
votes
4answers
3k views

Show that the ideal $ (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal

Show that the ideal $ I = (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal. My thoughts so far: Work by contradiction. Assume that $ I $ is principal, i.e. that it is generated ...
6
votes
3answers
1k views

What does this number theory statement mean?

I recently started studying number theory by myself and I am reading a book about number theory. There is one thing that I don't understand, the statement below: If $a,b \in \mathbb{Z}$, then there ...
5
votes
3answers
1k views

On the Importance of the Second and Fourth Isomorphism Theorems

I suppose I'd like to focus on the theorems for groups and rings, first of all. In particular, I'd rather not see anything about modules, simply because I don't feel I know enough about them. Anyway, ...
6
votes
4answers
1k views

Prove that in any GCD domain every irreducible element is prime

The proof of the following proposition is not completely clear to me. I get everything up until the bold part and I have a feeling some crucial steps are omitted, can anybody help clear this up? ...
1
vote
4answers
221 views

Prove $4+\sqrt{5}$ is prime in $\mathbb{Z}[\sqrt{5}]$

Prove $4+\sqrt{5}$ is prime in $\mathbb{Z}[\sqrt{5}]$. I already proved it's irreducible but have no idea what to do next since I dom't know much about $\mathbb{Z}[\sqrt{5}]$, if it is a UFD or not. ...
4
votes
3answers
505 views

If $\gcd(a,c)=1=\gcd(b,c)$, then $\gcd(ab,c)=1$ [duplicate]

As stated in the title, the problem to prove is Let $a,b,c \in \mathbb{Z}$. If $\gcd(a,c)=1=\gcd(b,c)$, then $\gcd(ab,c)=1$. I think I've proved it, but I would like a second opinion. Here goes: ...
2
votes
2answers
715 views

Can Euclid's Division Algorithm and/or Fundamental Theorem of Arithmetic implies this property of prime numbers

There is an exercise on page 44 of Amann's book Analysis, Vol I which stuck me so much. I quoted it here: Ex7: Let $p\in\mathbb{N}$ with $p>1.$ Prove that $p$ is a prime number if and only if, ...
2
votes
3answers
773 views

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.

Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some ...
3
votes
4answers
273 views

Does $\,x\equiv a\pmod{\! 2},\ x\equiv a\pmod{\! 5}\,\Rightarrow\,x\equiv a\pmod{\!10}?$

Problem: Find the units digit of $3^{100}$ using Fermat's Little Theorem (FLT). My Attempt: By FLT we have $$3^1\equiv 1\pmod2\Rightarrow 3^4\equiv1\pmod 2$$ and $$3^4\equiv 1\pmod 5.$$ Since $\gcd(2,...
-1
votes
1answer
577 views

Every maximal ideal is a prime ideal [duplicate]

The question formuled in the exam was exactly: ''Every maximal ideal is a prime ideal'' Maximal and prime ideals are defined for a commutative ring R, but the proof I have for maximal ideal $\...
4
votes
2answers
146 views

If a prime $p$ divides $n^2$ then it also divides $n$ - Is this proof correct?

I'm learning Real Analysis by myself and I wanted to prove that if a prime $p$ divides $n^2$ where $n$ is an integer, then $p$ divides $n$ itself. I saw that proving this is the same as saying that ...

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