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PostPosted: 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.


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PostPosted: 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...


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PostPosted: 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 \(\displaystyle{x\,,y\in I}\) ?


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PostPosted: 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! :clap2:


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PostPosted: 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?


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PostPosted: Tue Nov 24, 2015 3:06 pm 
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Since the ring is commutative, we get :

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


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PostPosted: 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"?


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PostPosted: 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.


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PostPosted: Tue Mar 15, 2016 4:31 pm 
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Additional question :

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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