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I can see that clearly if you take $\mathbb{Z}[x]/7\mathbb{Z}[x]$ and then quotient it by $(x^2 + 1)$ all the remaining elements will be of form $ax + b$ where $a$ can be 7 things and $b$ can be 7 things, so this "object" has 49 elements. I'm not sure how to go about showing that $\mathbb{Z}[x]/(7, x^2 + 1)$ is isomorphic to this "object" and furthermore, that this object is necessarily a field. Any help you could give would be much appreciated. Thanks in advance.

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2 Answers

up vote 6 down vote accepted

Hint 1: $X^2+1$ is irreducible in $\mathbb{Z}/7 \mathbb{Z} [X]$.

Hint 2: Find the Kernel and the image of the composition: $$\mathbb Z [X] \to \left(\mathbb{Z}/7 \mathbb{Z}\right) [X] \to \frac{\left(\mathbb{Z}/7 \mathbb{Z}\right) [X]}{\langle X^2+1 \rangle} \,.$$

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Fleshing out a little the already complete answer by N.S.: take any element $\,f(x)\in\Bbb Z[x]\,$ , reduce its coefficients modulo $\,7\,$ so that you get a map:

$$\Bbb Z[x]\to \Bbb F_7[x]\,\,,\,\,f(x)\to \overline f(x):=f(x)\pmod 7\,\,\,,\,\,\Bbb F_7:=\Bbb Z/7\Bbb Z$$

and now divide $\,\overline f(x)\,$ by $\,x^2+1\,$ in $\,\Bbb F_7[x]\,$ with residue:

$$\overline f(x)=g(x)(x^2+1)+r(x)\,\,,\,\,g(x),r(x)\in\Bbb F_7[x]\,\,\,,\,\,\deg r<2\,\,\,or\,\,\,r(x)=0$$

so that we can map $\,\overline f(x)\to r(x)+(x^2+1)\in K\,$ , from $\,\Bbb F_7[x]\to K:=\Bbb F_7[x]/(x^2+1)\,$

Finally, observe that $\,K\,$ is a field since $\,x^2+1\,$ is irreducible in $\,\Bbb F_7[x]\,$ since $\,-1\,$ is a square $\,\pmod p\Longleftrightarrow p=1\pmod 4\,$...

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