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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...

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

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!

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?

Since the ring is commutative, we get :

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

Additional question: Is statement (1) true if the word "prime" is replaced by "maximal"?

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.

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 \(\displaystyle{\left(R,+,\cdot\right)}\) .