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 Post subject: On Ring Theory (An Easy One) Posted: Sat Nov 21, 2015 4:48 pm
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1. Let $\displaystyle \phi : B \longrightarrow A$ be a ring homomorphism. If $\displaystyle \mathscr{p}$ is a prime ideal of $A$, then show that $\phi^{-1}(\mathscr{p})$ is a prime ideal of $B$.
2. Show that the nilpotents of a ring $R$ form an ideal.

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Mon Nov 23, 2015 2:40 pm
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Hi Nickos.

I suppose that you mean associative rings with unity.

1. Let $\displaystyle{P}$ be a prime ideal of $\displaystyle{A}$ .

Suppose that $\displaystyle{I_{1}\,,I_{2}}$ are two ideals of $\displaystyle{B}$, such that

$\displaystyle{I_1\,I_2\subseteq \phi^{-1}(P)}$ . Then, $\displaystyle{\phi(I_1,I_2)\subseteq \phi(\phi^{-1}(P))\subseteq P}$ .

Since $\displaystyle{\phi}$ is homomorphism, we get $\displaystyle{\phi(I_1)\,\phi(I_{2})\subseteq P}$ .

Now, $\displaystyle{\phi(I_{1})\,,\phi(I_{2})}$ are ideals of $\displaystyle{A}$ and $\displaystyle{P}$

is a prime ideal of $\displaystyle{A}$, so :

$\displaystyle{\phi(I_{1})\subseteq P}$ or $\displaystyle{\phi(I_{2})\subseteq P}$, that is :

$\displaystyle{I_{1}\subseteq \phi^{-1}(\phi(I_{1})\subseteq\phi^{-1}(P)}$ or

$\displaystyle{I_{2}\subseteq \phi^{-1}(\phi(I_{2})\subseteq\phi^{-1}(P)}$.

Therefore, $\displaystyle{\phi^{-1}(P)}$ is a prime ideal of $\displaystyle{B}$ .

To be continued...

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Mon Nov 23, 2015 3:02 pm
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2.

Let $\displaystyle{I}$ be the set of all the nilpotents elements of a commutative ring with unity.

Obviously, $\displaystyle{0\in I}$. If $\displaystyle{x\in I}$, then $\displaystyle{x^n=0}$

for some $\displaystyle{n\in\mathbb{N}}$ . Then, $\displaystyle{(-x)^n=(-1)^n\,x^n=0}$, so

$\displaystyle{-x\in I}$ . If $\displaystyle{r\in R}$, then $\displaystyle{(x\,r)^n=(r\,x)^n=r^n\,x^n=0}$,

so $\displaystyle{r\,x\,,x\,r\in I}$ .

What about $\displaystyle{x+y}$ if $\displaystyle{x\,,y\in I}$ ?

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Mon Nov 23, 2015 9:57 pm
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Papapetros Vaggelis wrote:
I suppose that you mean associative rings with unity.

Ιn fact, I had in mind that the rings are commutative (and associative, of course) with unity - influenced by my recent studying. Well done! Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Mon Nov 23, 2015 10:05 pm
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Papapetros Vaggelis wrote:
What about $\displaystyle{x+y}$ if $\displaystyle{x\,,y\in I}$ ?

Suppose that $\displaystyle x^{n} = 0$ and $\displaystyle y^{m} = 0$. If you expand the expression $\displaystyle (x+y)^{n+m-1}$, you will obtain $\displaystyle (x+y)^{n+m-1} =0$. Can you see why?

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Tue Nov 24, 2015 3:06 pm
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Since the ring is commutative, we get :

\displaystyle{\begin{aligned} (x+y)^{n+m-1}&=\sum_{k=0}^{n+m-1}\binom{n+m-1}{k}x^k\,y^{n+m-1-k}\\&=\sum_{k=0}^{n-1}\binom{n+m-1}{k}x^{k}\,(y^{m})\,y^{(n-1)-k}+\sum_{k=n}^{n+m-1}\binom{n+m-1}{k}x^{(k-n)}\,x^n\,y^{n+m-1-k}\\&=0\end{aligned}}

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Fri Dec 04, 2015 3:47 pm
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Additional question: Is statement (1) true if the word "prime" is replaced by "maximal"?

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Tue Mar 15, 2016 9:03 am
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Quote:
Is statement (1) true if the word "prime" is replaced by "maximal"?

No! For example, consider the natural inclusion $f \ \colon \mathbb{Z} \hookrightarrow \mathbb{Q}$. Then the zero ideal $0$ of $\mathbb{Q}$ is maximal, as $\mathbb{Q}$ is a field, while the inverse image $f^{-1}(0)$ of $0$ under $f$ is the zero ideal $0$ of $\mathbb{Z}$ which is just a prime ideal.

Top   Post subject: Re: On Ring Theory (An Easy One) Posted: Tue Mar 15, 2016 4:31 pm
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If $\displaystyle{\left(R,+,\cdot\right)}$ is a commutative ring with unity, then prove that

$\displaystyle{\sqrt{0}=\bigcap_{P\in \rm{Spec}(R)}P}$ ,

where,

$\displaystyle{\rm{Spec}(R)}$ is the set of all the prime ideals of $\displaystyle{\left(R,+,\cdot\right)}$ .

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