Write two distinct elements of $\mathbb Z[x] / \langle x-7, 13\rangle$.
Distinct elements of $\mathbb{Z}[x]/I$
3 Answers
Hint $\rm\ 1\not\equiv 0\pmod I\:$ else $\rm\:1\in I\ \Rightarrow\ 1\: =\: 13\:f(x) + (x-7)\:g(x)\ \Rightarrow\ 1 = 13\:f(7) \:$ in $\mathbb Z$
Remark $\ $ This shows that $\rm\:\mathbb Z\to \mathbb Z[x]\to \mathbb Z[x]/I\:$ has kernel $\rm\:13\!\:\mathbb Z.\:$ Further, this map is onto since $\rm\:f(x)\equiv f(7)\pmod{I}.\:$ Hence $\rm\:\mathbb Z[x]/I\:\!\cong\!\: \mathbb Z/13\:$ by the First Isomorphism Theorem.
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0@BillDubuqueMay I please ask you if this is correct? In the language of Artin: "kill" (x-7) in Z[x], which sends x to 7 in Z. Then "kill" 13 in Z (since no variable x in the constant 13) this in then isomorphic to Z/13Z. (Tried many times to correctly format this, but got timed out at 5 min. My apologies.) Thanks – 2012-03-26
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1@Andrew That's essentially the same as my remark, viz. $\rm\:mod\ I\!:\ f(x)\equiv (f(7)\ mod\ 13),\:$ which may be computed by first reducing $\rm\:f(x)\ mod\ (x-7)\: =\: f(7),\:$ then reducing $\rm\:f(7)\ mod\ 13.\:$ Thus the map $\rm\:f(x)\to f(7) + 13\:\mathbb Z\:$ yields $\rm\:\mathbb Z[x]/I\:\!\cong\!\: \mathbb Z/13$. – 2012-03-27
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0@BillDubuqueThanks – 2012-03-27
What is $x$ in this ring? Once you know that, you should have a good idea what the ring looks like, and from there the problem is easy.
Here's a cheap answer (that you should NOT use as your solution): The problem asks for two distinct elements of the ring. Therefore, (for completely non-mathematical reasons) we assume that the ring has 2 distinct elements. We know $1$ is in the ring, and so is $0$, and $1 \neq 0$ in all rings with unit except the ring with one element.
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0x-7+13=0 => x= -6 is this correct? – 2012-03-26
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0Yes, this will do. There's an even simpler way of representing $x$ as an integer, though. – 2012-03-26
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0Sir, please teach me the simpler way. – 2012-03-26
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0$x-7=0$, so $x=7$. Now, what is this ring isomorphic to? – 2012-03-26
Let $\pi : \mathbb{Z}[x] \to \mathbb{Z}[x]/I$ be the canonical projection. We claim that $\pi(1)$ and $\pi(2)$ are distinct elements in $\mathbb{Z}[x]/I$. Otherwise, $2 - 1 = 1 \in I$, and $I$ would be the unit ideal. Now, show that $I$ is not, in fact, the unit ideal.