- If \( \mathfrak{b} \) is an ideal of \( B \), then \( \displaystyle f^{-1}(\mathfrak{b}) \) is an ideal of \( A \) (called the contraction of \( \mathfrak{b} \)). Moreover, If \( \mathfrak{b} \) is a prime ideal of \( B \), then \( \displaystyle f^{-1}(\mathfrak{b}) \) is a prime ideal of \( A \).
- If \( \mathfrak{a} \) is an ideal of \( A \), then \( \displaystyle f(\mathfrak{a}) \) need not be an ideal of \( B \). Moreover, even if \( \mathfrak{a} \) is a prime ideal of \( A \), then \( \displaystyle f(\mathfrak{a}) \) need not be a prime ideal of \( B \).
Basic Ring Theory - 11
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Basic Ring Theory - 11
Let \( A \) and \( B \) be commutative rings with unity \( 1_{A} \) and \( 1_{B} \), respectively, and let \( \displaystyle f \ \colon A \longrightarrow B \) be a ring homomorphism. Show that
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Re: Basic Ring Theory - 11
i. Let \(\displaystyle{b}\) be an ideal of \(\displaystyle{B}\). Since,
\(\displaystyle{f(0_{A})=0_{B}\in b}\), we have that \(\displaystyle{0_{A}\in f^{-1}(b)}\).
Consider \(\displaystyle{x\,,y\in f^{-1}(b)}\) and \(\displaystyle{a\in A}\).
The function \(\displaystyle{f}\) is a ring homomorphism, so,
\(\displaystyle{f(x)-f(y)=f(x-y)\in b\implies x-y\in f^{-1}(b)}\) (because \(\displaystyle{f(x)\,,f(y)\in b}\) ) and
\(\displaystyle{f(x\,a)=f(a\,x)=f(a)\,f(x)\in b\implies x\,a=a\,x\in f^{-1}(b)}\)
(because \(\displaystyle{f(a)\in B\,,f(x)\in b}\) and \(\displaystyle{b}\) is an ideal of \(\displaystyle{B}\)).
We conclude that \(\displaystyle{f^{-1}(b)}\) is an ideal of \(\displaystyle{A}\).
Suppose now that the ideal \(\displaystyle{b}\) is also a prime ideal of \(\displaystyle{B}\).
Let \(\displaystyle{x\,,y\in A}\) such that \(\displaystyle{x\,y\in f^{-1}(b)}\). Then,
\(\displaystyle{f(x\,y)\in b\implies f(x)\,f(y)\in b\implies f(x)\in b\,\lor f(y)\in b\implies x\in f^{-1}(b)\,\lor y\in f^{-1}(b)}\).
Therefore, \(\displaystyle{f^{-1}(b)}\) is a prime ideal of \(\displaystyle{A}\).
ii.
Consider the commutative rings \(\displaystyle{A=\left(\mathbb{Z},+,\cdot\right)\,,B=\left(\mathbb{Q},+,\cdot\right)}\)
and the ring homomorphism \(\displaystyle{i:\mathbb{Z}\to \mathbb{Q}\,,i(x)=x}\) .
If \(\displaystyle{a=\mathbb{Z}}\), then \(\displaystyle{a}\) is an ideal of \(\displaystyle{A}\) but
\(\displaystyle{f(a)=f(\mathbb{Z})=\mathbb{Z}}\) is not an ideal of \(\displaystyle{B}\), since
\(\displaystyle{\dfrac{1}{2}\in\mathbb{Q}\,,1\in\mathbb{Z}}\) and \(\displaystyle{\dfrac{1}{2}\cdot 1=\dfrac{1}{2}\notin \mathbb{Z}}\).
To be continued...
\(\displaystyle{f(0_{A})=0_{B}\in b}\), we have that \(\displaystyle{0_{A}\in f^{-1}(b)}\).
Consider \(\displaystyle{x\,,y\in f^{-1}(b)}\) and \(\displaystyle{a\in A}\).
The function \(\displaystyle{f}\) is a ring homomorphism, so,
\(\displaystyle{f(x)-f(y)=f(x-y)\in b\implies x-y\in f^{-1}(b)}\) (because \(\displaystyle{f(x)\,,f(y)\in b}\) ) and
\(\displaystyle{f(x\,a)=f(a\,x)=f(a)\,f(x)\in b\implies x\,a=a\,x\in f^{-1}(b)}\)
(because \(\displaystyle{f(a)\in B\,,f(x)\in b}\) and \(\displaystyle{b}\) is an ideal of \(\displaystyle{B}\)).
We conclude that \(\displaystyle{f^{-1}(b)}\) is an ideal of \(\displaystyle{A}\).
Suppose now that the ideal \(\displaystyle{b}\) is also a prime ideal of \(\displaystyle{B}\).
Let \(\displaystyle{x\,,y\in A}\) such that \(\displaystyle{x\,y\in f^{-1}(b)}\). Then,
\(\displaystyle{f(x\,y)\in b\implies f(x)\,f(y)\in b\implies f(x)\in b\,\lor f(y)\in b\implies x\in f^{-1}(b)\,\lor y\in f^{-1}(b)}\).
Therefore, \(\displaystyle{f^{-1}(b)}\) is a prime ideal of \(\displaystyle{A}\).
ii.
Consider the commutative rings \(\displaystyle{A=\left(\mathbb{Z},+,\cdot\right)\,,B=\left(\mathbb{Q},+,\cdot\right)}\)
and the ring homomorphism \(\displaystyle{i:\mathbb{Z}\to \mathbb{Q}\,,i(x)=x}\) .
If \(\displaystyle{a=\mathbb{Z}}\), then \(\displaystyle{a}\) is an ideal of \(\displaystyle{A}\) but
\(\displaystyle{f(a)=f(\mathbb{Z})=\mathbb{Z}}\) is not an ideal of \(\displaystyle{B}\), since
\(\displaystyle{\dfrac{1}{2}\in\mathbb{Q}\,,1\in\mathbb{Z}}\) and \(\displaystyle{\dfrac{1}{2}\cdot 1=\dfrac{1}{2}\notin \mathbb{Z}}\).
To be continued...
Last edited by Papapetros Vaggelis on Sat Apr 02, 2016 12:28 pm, edited 1 time in total.
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Re: Basic Ring Theory - 11
Consider now the map
\(\displaystyle{\Phi:\mathbb{R}[x]\to \mathbb{R}\,\,,\Phi\,\left(\sum_{i=0}^{n}a_{i}\,x^{i}\right)=a_0}\).
Obviously the map \(\displaystyle{\Phi}\) is a ring homomorphism.
If \(\displaystyle{P=\langle{x^2+1\rangle}}\), then \(\displaystyle{P}\) is a prime ideal of \(\displaystyle{\mathbb{R}[x]}\) .
We have that \(\displaystyle{\Phi(P)\subseteq \mathbb{R}}\).
Let \(\displaystyle{a\in\mathbb{R}}\). Then, \(\displaystyle{f(x)=a\,x^2+a=a(x^2+1)\in P}\) and
\(\displaystyle{\Phi(f(x))=a\implies a\in\Phi(P)}\). Therefore,
\(\displaystyle{\Phi(P)=\mathbb{R}}\) which is not a prime ideal of \(\displaystyle{\mathbb{R}}\).
\(\displaystyle{\Phi:\mathbb{R}[x]\to \mathbb{R}\,\,,\Phi\,\left(\sum_{i=0}^{n}a_{i}\,x^{i}\right)=a_0}\).
Obviously the map \(\displaystyle{\Phi}\) is a ring homomorphism.
If \(\displaystyle{P=\langle{x^2+1\rangle}}\), then \(\displaystyle{P}\) is a prime ideal of \(\displaystyle{\mathbb{R}[x]}\) .
We have that \(\displaystyle{\Phi(P)\subseteq \mathbb{R}}\).
Let \(\displaystyle{a\in\mathbb{R}}\). Then, \(\displaystyle{f(x)=a\,x^2+a=a(x^2+1)\in P}\) and
\(\displaystyle{\Phi(f(x))=a\implies a\in\Phi(P)}\). Therefore,
\(\displaystyle{\Phi(P)=\mathbb{R}}\) which is not a prime ideal of \(\displaystyle{\mathbb{R}}\).
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