$\mathbb{R}[x]/\langle X^2+1 \rangle$, kernel and the first isomorphism theorem

Question

Background

Proposition 4.33 In the ring $K[X]$ of polynomials over a field $K$ every ideal is principal. If $J$ is a non-zero ideal of $K[X]$ and $g$ in $J$ is a non-zero polynomial of minimal degree among the non-zero polynomials of $J$, then $g$ generates $J$. Any two generators of $J$ are constant multiples of each other.

The substitutino map $\phi_1:\mathbb{R}[X]\to \mathbb{C}$ given by $f(X)\mapsto f(i)$ is surjectibe. Indeed, if $a,b\in \mathbb{R}$, then $\phi_1(a+bX)=a+bi$, which covers all complex numbers. According to Proposition 4.33, $\text{ker }\phi_1$ is a pirincipal ideal generated by a polynomial in $\text{ker }\phi_1$ of minimal degree. Well, $X^2 +1$ is such a polynomial, since $\phi_1(X^2 +1)=i^2 +1=0$ and since no polynomial of lower degree linear or constant, lies in $\text{ker }\phi_1$. By the first isomorphism theorem,

$$\mathbb{C}\cong\mathbb{R}[x]/\langle X^2+1 \rangle$$.

Question

I would like to know suppose I am simply given the map $\phi_1:\mathbb{R}[X]\to \mathbb{C}$ defined by $\phi_1(a+bX)=a+bi$, then how would i determine the kernel of $\phi_1$ as $\langle X^2+1 \rangle$ without resorting to Proposition 4.33 above. I mean how does it go from $a+bX$ to $X^2 +1$ in the domain of $\phi_1$. In the end, we want $\phi_1(a+bX)=a+bi=0$, and the only information we have is $x\mapsto i$. But again the domain of $\phi_1$ seem to be asking for a polynomial of specific degree, but it starts with a linear polynomial.

Thank you in advance.

Answer 1

Let $z = a + bi$. The map $ \phi_1$ is simply evaluating a polynomial at $x = z$ it therefore follows that the kernel of $\phi_1$ is exactly the set of polynomials $f$ such that $f(z) = 0$. We deduce that $\overline z$ is also a root of $f \in \ker \phi_1$ by using

If $z \in \mathbb C$ is a root of a real polynomial then so is $\overline z \in \mathbb C$.

Therefore each polynomial $f$ in the kernel is divisible by $(X-z)$ and $(X- \overline z)$ when viewed as a polynomial in $\mathbb C [X]$which means that $f$ is divisible by

$$(X - \overline z)(X- z) = X^2 -( z + \overline z)X + \overline zz = X^2-2a X + \vert z \vert^2$$ in $\mathbb C[X].$

Since $f$ is divisible by $X^2-2a X + \vert z \vert^2$ in $\mathbb C[X]$ we can use the long division algorithm to find $q \in \mathbb C[X]$ such that

$$ f(X) = q(X) (X^2 - 2aX + \vert z \vert^2).$$

since $f$ and $X^2 - 2aX + \vert z \vert^2$ all have real coefficients the long division algorithm will give a $q$ with real coefficients. This proves that $f$ is divisible by $X^2 - 2aX + \vert z \vert^2$ in $\mathbb{R}[X]$.

Therefore each polynomial in the kernel is divisible by $X^2 - 2aX + \vert z \vert^2$

It follows that

$$ \ker \phi_1 = \langle X^2 - 2a X + \vert z \vert^2 \rangle $$

Answer 2

I was hoping to get an answer in this similar to what jgon stated in this post, via some sort of algebraic derivation

Below I explain how to do it that way. First I clarify that way - highlighting the use of the norm (multiplying by the conjugate) to $\rm\color{#90F}{rationalize}$ the denominator, which simplifies division by complex integers to division by integers. OP works the same way, except here we $\rm\color{#90F}{realize}$ the denominator, which reduces division by complex polynomials to real polynomials. More generally see the method of $\rm\color{#90F}{simpler\ multiples}$.

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$\bbox[4px,border:1px solid #c00]{A=\Bbb Z[i]/(3\!+\!i)\equiv \Bbb Z/10}\ $ $\ \ \Bbb Z \overset{h}\to A\,\color{#0a0}{ \ {\rm is\ onto}}\,$ by $\!\bmod\, 3\!+\!i\!:\ \, i\equiv -3\,\Rightarrow\, a\!+\!bi\equiv a\!-\!3b\color{#0a0}{\in\Bbb Z}$

$\color{#c00}{I := \ker h = 10\,\Bbb Z}\ $ follows immediately by $\,\rm\color{#90F}{rationalizing}\,$ a denominator

$$ n\in I\!\iff\! 3\!+\!i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\!\iff\! \dfrac{n}{3\!+\!i} \color{#90f}{\overset{\large \rm\ rat}=} \dfrac{n(3\!-\!i)}{10}=\frac{3n}{10}\!-\!\color{#c00}{\frac{n}{10}}\:\!i\in\Bbb Z[i]\!\iff\! \color{#c00}{10\mid n}\qquad\ $$

Thus $\, \color{#0a0}{A = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{10\,\Bbb Z}\ $ by the First Isomorphism Theorem.

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$\bbox[4px,border:1px solid #c00]{\Bbb R[x]/(x^2\!+\!1)\equiv \Bbb C}\ $ $\ \ \Bbb R[x]\overset{h}\to \Bbb C,\ f(x)\mapsto f(i)\,\color{#0a0}{ \ {\rm is\ onto}}\,$ by $\,h(a\!+\!bx) = a\!+\!bi$.

$\color{#c00}{I := \ker h = (x^2\!+\!1)\,\Bbb R[x]}\ $ follows immediately by $\,\rm\color{#90F}{realizing}\,$ a denominator

$\ f\!\in\! I\!\iff\! x\!-\!i\mid f\ \, {\rm in}\, \ \Bbb C[x]\!\iff\! \dfrac{f}{x\!-\!i}\color{#90f}{\overset{\!\large \rm\ rea}=} \dfrac{(x+i)f\!}{x^2+1}=\underbrace{\dfrac{xf}{x^2+1}+\color{#c00}{\dfrac{f}{x^2+1}}\:\!i\:\!\in\:\!\Bbb C[x]}_{\textstyle\ \ \ \ \iff \color{#c00}{{x^2\!+\!1}\mid f\ \,{\rm in}\,\ \Bbb R[x]} }$

Thus $\, \color{#0a0}{\Bbb C = {\rm Im}\ h} \,\cong\, \Bbb R[x]/\color{#c00}{\ker h} \,=\, \Bbb R[x]/\color{#c00}{(x^2\!+\!1)}\ $ by the First Isomorphism Theorem.

Answer 3

Clearly you have $\langle x^2+1\rangle \subseteq \ker\phi_1$, since $x^2+1\in\ker\phi_1$ and $\phi_1$ is a map of rings.
Conversly, assume that $f\in\ker\phi_1$

Then we have: $\phi_1(f)=f(i)=0$.
So $i$ is a root of the polynomial $f$.
Since $f$ is a polynomial with real coefficients, it must be the case that $\bar{i}=-i$ is also a root of $f$. (Do you see why?)

As $i,-i$ are roots of $f$, there exists a polynomial $g\in\mathbb{R}[x]$ such that: $$f(x)=(x-i)(x+i)g(x)=(x^2+1)g(x)$$ But this is equivalent to $f\in\langle x^2+1\rangle$.

Since the $f\in\ker\phi_1$ we started with was arbitrary, we conclude that $\ker\phi_1\subseteq\langle x^2+1\rangle$, proving the inverse inclusion, giving us the wanted equality $\langle x^2+1\rangle=\ker\phi_1$