Understanding properties of quotient rings

Question

We have begun learning quotient rings in my Algebra course, but I am still confused by some of the theorems and properties of quotient rings.

We have a theorem:

$R$ is a ring, and $I\subset R$ is an ideal. Then there is a unique multiplication on $R/I$ making it into a ring such that $$\pi:R\rightarrow R/I$$ defined by $a\mapsto a+I$ is a ring homomorphism with $\ker \pi = I$.

I am confused why $\ker \pi = I$ is true.

If we take $R=\mathbb{Z}[x]$ and $I=(x^{2}+7)$, then we have $$\phi(x^{2}+7) = \phi(x^{2})+\phi(7) = \phi(x)^{2}+\phi(7) = \phi(x)^{2}+7+k(x^{2}+7) =0$$ I am not understanding why $\phi(x^{2}+7)=0$ must hold?

I am working on a problem where I must show that two quotient rings are not isomorphic, and I am struggling because I am still not quite understanding the properties of quotient rings other than the fact that it is an additive coset of an ideal $I$ in $R$.

Any insight is appreciated!

Answer 1

the ideal of $\mathbb{Z}[x]$ generated by the polynomial $x^2+7$ is denoted $I=(x^{2}+7)$ and its elements i.e. the equivalence classes of the quotient $\mathbb{Z}[x]/I$ are denoted $$\overline{f(x)}=f(x)+I\equiv f(x)+(x^2+7)$$ Intuitively these elements can be thought of as being obtained by setting $x^2+7=0$ in $f(x)$. So the projection $$\phi:\mathbb{Z}[x]\rightarrow \mathbb{Z}[x]/(x^2+7)$$ given by $\phi(f)=\bar{f}$ consists of performing the euclidean division of $f(x)$ with $x^2+7$ and keeping the remainder: $$f(x)=(x^2+7)q(x)+r(x)$$

and $$\phi\big(f(x)\big)=\overline{f(x)}=r(x)$$ Concluding: $x^2+7$ and its polynomial multiples are identically zero in the quotient ring $\mathbb{Z}[x]/(x^2+7)$.

An immediate consequence of the above is that the notation $\overline{f(x)}$ collectively identifies all integer polynomials $f(x)$, $g(x)$ which differ by a polynomial multiple of $x^2+7$: $$f(x)-g(x)=k(x)(x^2+7)\Leftrightarrow \overline{f(x)}=\overline{g(x)}$$ where $k(x)\in\mathbb{Z}[x]$.

Answer 2

Just apply the definitions. The set $R/I$ consists of the equivalence classes under the equivalence relation $\sim$ on $R$ defined by $a\sim b$ to stand for $a-b\in I$.

It happens that the equivalence classes can be represented as the sets $a+I=\{a+c:c\in I\}$, where $a+I$ is indeed the equivalence class containing $a$. On the set $R/I$ we can define operations by $$ (a+I)+(b+I)=(a+b)+I,\quad (a+I)(b+I)=ab+I $$ If these are well defined (that is, they don't depend on the representative of the equivalence classes), then clearly the map $\pi\colon R\to R/I$ is a ring homomorphism.

The fact that the operations are well defined stems from $I$ being an ideal: if $a\sim a'$ and $b\sim b'$, that is, $a-a'\in I$ and $b-b'\in I$, then $$ (a+b)-(a'+b')=(a-a')+(b-b')\in I $$ so $(a+b)+I=(a'+b')+I$ and $$ ab-a'b'=ab-a'b+a'b-a'b'=(a-a')b+a'(b-b')\in I $$ so $ab+I=a'b'+I$ and we are done.

The zero of $R/I$ is clearly $0+I$, so the kernel of $\pi$ is $$ \ker\pi=\{a\in R:a+I=0+I\}=I $$

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Now apply this to your specific case: $I=(x^2+7)$, an ideal of $\mathbb{Z}[x]$. If $\phi\colon\mathbb{Z}[x]\to\mathbb{Z}[x]/I$ is the canonical projection (denoted above by $\pi$), you have $$ \phi(x^2+7)=(x^2+7)+I=0+I $$ because $x^2+7\in I$. As a consequence, $$ 0+I=\phi(x^2+7)=\phi(x^2)+\phi(7)=\phi(x)^2+\phi(7) $$ so, if $\xi=\phi(x)$, we can write $\phi(7)=-\xi^2$.

In this case the restriction of $\phi$ to $\mathbb{Z}$ is injective: indeed, a nonzero integer $n$ is not a multiple of $x^2+7$, so $\phi(n)\ne0+I$. It's customary in this case to identify integers with their image in $\mathbb{Z}[x]/I$, so the above relation can be written as $7=-\xi^2$. In the quotient ring (which can be regarded as an extension of $\mathbb{Z}$) there's an element $\xi$ such that $\xi^2=-7$.

Moreover, any element of $\mathbb{Z}[x]/I$ can be written in a unique way as $a+b\xi$, where $a,b\in\mathbb{Z}$ (provided we use the identification above).