Integrate $F(z)=e^{iz}/(z^2+1)$ around the real segment from $-R$ to $R$ and the counterclockwise upper semicircle, with $R>1$. The only enclosed pole is $i$, whose residue is $e^{-1}/(2i)$. On the semicircle, $|e^{iz}|\le 1$ and $|z^2+1|\ge R^2-1$, so its integral has magnitude at most $\pi R/(R^2-1)\to 0$.

The residue theorem thus gives ${\int }_{-\infty }^{\infty }{e}^{ix}/(x^2+1)dx=\pi e^{-1}$. The line integral converges absolutely, so taking its real part is justified:

The inverse image $I={\theta }^{-1}(J)$ is an ideal, by preservation of differences and multiplication. It is proper since $\theta (1)=1\notin J$. If $ab\in I$, then $\theta (a)\theta (b)\in J$; primality of $J$ gives $\theta (a)\in J$ or $\theta (b)\in J$, hence $a\in I$ or $b\in I$. Therefore the inverse image is prime.

For maximality, take the inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ and $J=(0)$. This is a maximal ideal of the field $\mathbb{Q}$, but its inverse image is $(0)$ in $\mathbb{Z}$, strictly contained in the proper ideal $2\mathbb{Z}$, so it is not maximal. Preservation of identity matters for proper inverse images: without it the zero homomorphism could pull a proper ideal back to the whole ring.

Use the standard definition that a prime ideal is proper, in addition to the product condition. The quotient operations areIts zero and identity are $I$ and $1+I$, which differ because $I$ is proper. If $(r+I)(s+I)=I$, then $rs\in I$, so primality gives $r\in I$ or $s\in I$. Thus one factor is zero. The commutative quotient ring has no nonzero zero divisors and is therefore an integral domain.

The printed definition omits properness. If that omission were read literally, $I=R$ would satisfy its product condition, but $R/I$ would be the zero ring, which is not an integral domain under the standard convention $1\ne 0$. Properness is the necessary qualification, retained explicitly in this proof.

An ideal is an additive subgroup $I\subseteq R$ with $rI\subseteq I$ for all $r\in R$. A ring homomorphism preserves addition and multiplication; for the rings with identity here use the standard unital convention $\theta (1_R)=1_S$. If $a,b\in \ker \theta$, then $\theta (a-b)=0$, and if $r\in R$ then $\theta (ra)=\theta (r)\theta (a)=0$. Thus the kernel is an ideal.

A maximal ideal is a proper ideal $I\ne R$ such that an ideal $J$ satisfying $I\subseteq J\subseteq R$ is either $I$ or $R$.

Assume condition (iii), and factor $m={\prod }_j{f}_j^{e_j}$. If some $e_j\ge 2$, the primary decomposition theorem gives the nonzero summand $V_j=\ker f_j(T{)}^{e_j}$ on which the minimal polynomial is exactly $f_j^{e_j}$. The operator $f_j(T)$ cannot vanish on this whole summand, or its restriction would have minimal polynomial dividing $f_j$. Choose $v\in V_j$ with $f_j(T)v\ne 0$. Thencontradicting (iii) with $p=f_j$ and $n=e_j$. Hence every $e_j=1$, proving (iii) implies (ii).

Suppose $m$ has no repeated irreducible factor. For every polynomial $p$ and integer $n\ge 1$, its monic greatest common divisors satisfy $\gcd (p^n,m)=\gcd (p,m)$, since raising $p$ to a power changes no membership of an irreducible factor of the square-free $m$. This is the square-free minimal polynomial criterion for polynomial kernel stabilization. The kernel identity proved above givesThus (ii) implies (iii) over every field.

Over $\mathbb{C}$, the distinct irreducible factors of $m$ are linear. The primary decomposition theorem therefore reads $V={⨁}_j\ker (T-{\lambda }_{j}I)$, a direct sum of eigenspaces. Taking a basis in each summand gives an eigenbasis, proving (ii) implies (i) over the complex numbers.

Over $\mathbb{R}$, take $T=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Its minimal polynomial is $X^2+1$, which is an irreducible polynomial over $\mathbb{R}$ and has no repeated factor. But it has no real eigenvalues, so no real eigenbasis. This is the required counterexample. Over a general field, the phrase “no repeated factors” here refers to irreducible factors in $K[X]$, rather than a prior assumption that those factors split.

If a basis consists of eigenvectors, let ${\lambda }_1,\dots ,{\lambda }_r\in K$ be the distinct eigenvalues. A polynomial annihilates the diagonal action exactly when it vanishes at each ${\lambda }_j$. Consequently the minimal polynomial is ${\prod }_j(X-{\lambda }_j)$, a product of distinct linear factors. It therefore has no repeated irreducible factor, proving (i) implies (ii).