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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Generalizations of the Notion of Primes

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Generalizations of the Notion of Primes J
Exercise 0.7
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Thayaden
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Y by felixgotti
If $R$ is an intergral domain then $R[x]$ is too,
Let $f(x),g(x)\in R[x]$ such that they are non zero, denote
$$f(x)=\sum_{i=0}^{n}{a_{i}x^{i}}$$$$g(x)=\sum_{j=0}^{m}{b_{j}x^{j}}$$Taking $f(x)g(x)$ let $c_{m+n}=a_{n}b_{m}$ be the leading coefficient of $f(x)g(x)$. Notice that $a_n,b_m\neq 0$ thus $c_{m+n}\neq0$ so in short for $p(x)\in R[x]$ the leading coefficient of $p(x)$ cannot be $0$ thus $R[x]$ must be an integral domain.
Going the other way say $R[x]$ is an integral domain,
For the sake of contradiction let $ab=0$ such that $a,b\in R$ and $a,b\neq0$, let $f(x)=a, g(x)=b\in R[x]$ taking there product,
$$f(x)g(x)=ab=0$$This contradicts our previous statement thus $R$ is an integral domain!
Finally iff $R$ is an integral domain then $R[x]$ must also be an integral domain $\blacksquare$
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Exercise 0.7
Thayaden   0
Oct 2, 2024
If $R$ is an intergral domain then $R[x]$ is too,
Let $f(x),g(x)\in R[x]$ such that they are non zero, denote
$$f(x)=\sum_{i=0}^{n}{a_{i}x^{i}}$$$$g(x)=\sum_{j=0}^{m}{b_{j}x^{j}}$$Taking $f(x)g(x)$ let $c_{m+n}=a_{n}b_{m}$ be the leading coefficient of $f(x)g(x)$. Notice that $a_n,b_m\neq 0$ thus $c_{m+n}\neq0$ so in short for $p(x)\in R[x]$ the leading coefficient of $p(x)$ cannot be $0$ thus $R[x]$ must be an integral domain.
Going the other way say $R[x]$ is an integral domain,
For the sake of contradiction let $ab=0$ such that $a,b\in R$ and $a,b\neq0$, let $f(x)=a, g(x)=b\in R[x]$ taking there product,
$$f(x)g(x)=ab=0$$This contradicts our previous statement thus $R$ is an integral domain!
Finally iff $R$ is an integral domain then $R[x]$ must also be an integral domain $\blacksquare$
0 replies
Thayaden
Oct 2, 2024
0 replies
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Exercise 0.7
hkoo   5
N Jun 27, 2024 by felixgotti
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

Problem. Let $R$ be a commutative ring. Prove that $R$ is an integral domain iff $R[x]$ is an integral domain.

(1) only if
if $R[x]$ is not an integral domain, there exist non-zero $f,g \in R[x]$ s.t
\[
b_m \neq 0,c_n \neq 0, f\cdot g=\bigg( \sum_{i=0}^m b_i x^i \bigg) \cdot \bigg( \sum_{i=0}^n c_i x^i \bigg) := \sum_{i=0}^{m+n} d_i x^i=0.
\]
It implies that $d_i = 0 for i=0,...,m+n$. So $d_{m+n}=b_m*c_n=0$. It's contradiction to fact that $R$ is integral domain. therefore $R[x]$ is integral domain.

(2) if
for $f \in R[x]$ s.t degree of $f$= 0, it's element of $R$. Since R is subset of R[x], R is also integral domain.
Where I'm stuck:

because it is easy to show that $\deg \, (fg) := \deg \, f + \deg \, g$ for all nonzero $f,g \in R[x]$, i wouldn't write a proof.
<Describe what's confusing you, or what your question is here!>
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hkoo
May 26, 2024
felixgotti
Jun 27, 2024
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Exercise 0.7
Thayaden   0
Oct 2, 2024
If $R$ is an intergral domain then $R[x]$ is too,
Let $f(x),g(x)\in R[x]$ such that they are non zero, denote
$$f(x)=\sum_{i=0}^{n}{a_{i}x^{i}}$$$$g(x)=\sum_{j=0}^{m}{b_{j}x^{j}}$$Taking $f(x)g(x)$ let $c_{m+n}=a_{n}b_{m}$ be the leading coefficient of $f(x)g(x)$. Notice that $a_n,b_m\neq 0$ thus $c_{m+n}\neq0$ so in short for $p(x)\in R[x]$ the leading coefficient of $p(x)$ cannot be $0$ thus $R[x]$ must be an integral domain.
Going the other way say $R[x]$ is an integral domain,
For the sake of contradiction let $ab=0$ such that $a,b\in R$ and $a,b\neq0$, let $f(x)=a, g(x)=b\in R[x]$ taking there product,
$$f(x)g(x)=ab=0$$This contradicts our previous statement thus $R$ is an integral domain!
Finally iff $R$ is an integral domain then $R[x]$ must also be an integral domain $\blacksquare$
0 replies
Thayaden
Oct 2, 2024
0 replies
J